When a ball is projected horizontally, what is its initial vertical velocity?

Table of Contents Show

Initial Velocity and Launch Angle
Practice Question 1
Changes in Vertical Velocity
Practice Question 2

initial velocity
launch angle
maximum height
time of flight
final velocity
launch height
horizontal range of the projectile

Initial Velocity and Launch Angle

All objects at the beginning of their projectile motion must possess a non-zero initial velocity.
The initial velocity can always analysed as and resolved into two components: horizontal and vertical velocities. This is done by constructing a right-angled triangle from vectors.

Conventions for calculations
- Initial velocity is typically written as u
- Initial vertical velocity is typically written as uy – subscript y is used to represent the vertical rectilinear motion
- Initial horizontal velocity is typically written as ux – subscript x is used to represent the horizontal rectilinear motion.

The relationship between initial velocity, initial horizontal and vertical velocity can always be represented by the right-angled triangle with q (as shown in the diagram) is the angle at which the projectile leaves the horizontal plane (usually the ground). This angle is the launch angle.

Using trigonometry, initial horizontal and initial vertical velocities can be expressed in terms of the initial velocity.

Initial horizontal velocity:

`u_x/u=costheta`

Initial vertical velocity:

`u_y/u=sintheta`

The relationship between initial vertical and horizontal velocity is described by:

`(u_y)/(u_x)=tantheta`

Since the triangle is right-angled, the three vectors’ relationship can also be summarised by Pythagoras’s theorem.

`u^2=(u_x)^2+(u_y)^2`

`u=sqrt((u_x)^2+(u_y)^2)`

The initial velocity can be negative because the initial direction of a projectile can also be downwards as shown below. Situations in which this type of initial velocity occurs will be explored and clarified in practice questions later.

Practice Question 1

A projectile is launched at 60 ms-1 at an elevation of 300. Find its initial horizontal and vertical velocities.

Construct a right-angled triangle from vectors:

Initial horizontal velocity:

`u_x=60cos30^o`

Initial vertical velocity:

`u_y=60sin30^o`

Initial vertical velocity changes throughout projectile motion. Its magnitude decreases when a object travels upwards and increases when it travels downwards.
Initial horizontal velocity remains constant and does not change.

Changes in Vertical Velocity

Motion in the vertical axis can be modelled using rectilinear equations. In contrast, motion in the horizontal axis does not require these equations because horizontal acceleration is zero.

s represents the vertical displacement, ay represents the acceleration in the vertical axis (gravity), uy and vy are the initial vertical velocity and velocity after time t of the object during its projectile motion.

Practice Question 2

The initial vertical velocity of an object during projectile motion is 15 ms-1. The launch angle is `30^o`.

(a) Calculate the initial horizontal velocity.

(b) Calculate the vertical velocity and the instantaneous velocity 2 seconds after the object’s launch.

The vertical velocity can be modelled by the following equation:

`v_y=u_y+a_yt`

Since, acceleration due to gravity is acting against the direction of the object, a negative sign must be placed in front of 9.8.

The instantaneous velocity can be determined using Pythagoras’s theorem:

`v=sqrt((25.98)^2+(-4.6)^2)`

Since velocity is a vector quantity, a direction is needed. The direction is typically indicated by the angle relative to the horizontal.

Using trigonometry:

Therefore, the instantaneous velocity of the object 2 seconds after launch is 26.4 ms-1 10.0° relative to the horizontal axis.

Construction of a right-angled triangle using velocity vectors can be done at any point during an object’s projectile motion.
- The horizontal velocity (horizontal vector) remains constant and does not change in length.
- The angle, vertical velocity (vertical vector) and the instantaneous velocity (hypotenuse) change with time.

Previous section: Introduction to Projectile Motion

Next section: Full-flight Projectile Motion - Maximum Height, Time of Flight and Range

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In this explainer, we will learn how to solve problems about projecting bodies horizontally from a point above the ground.

When a particle is projected horizontally, its initial velocity in the vertical direction is zero. As long as the particle is moving freely under gravity, then gravity is the only force acting upon it. This means that its horizontal acceleration is zero (so its velocity in the horizontal direction is constant) and that it has a constant vertical acceleration of 𝑔 downward.

We recall the equations of motion.

If a particle has initial velocity 𝑢 and constant acceleration 𝑎, then its displacement 𝑠 at time 𝑡 is given by 𝑠=𝑢𝑡+12𝑎𝑡 or 𝑠=𝑢+𝑣2𝑡.

Its velocity 𝑣 at time 𝑡 is given by 𝑣=𝑢+𝑎𝑡.

For a particle projected horizontally, these equations take a simpler form. We can analyze the horizontal and vertical components of the particle’s motion separately. A horizontally projected particle moving freely under gravity has no horizontal acceleration. Therefore, its horizontal velocity does not change, so in the horizontal direction, 𝑣=𝑢 and 𝑠=𝑢𝑡.

On the other hand, a particle projected horizontally has zero initial vertical velocity and accelerates downward because of gravity, so in the vertical direction, 𝑣=𝑔𝑡 (notice that 𝑣 and 𝑔 have the same sign here as they are both pointing downward) and 𝑠=12𝑔𝑡 (similarly, 𝑠 and 𝑔 have the same sign here).

A particle was projected horizontally from a point 42 m above the ground at 32 m/s. Find, to one decimal place, the time it took the particle to reach the ground. Take 𝑔=9.8/ms.

Answer

First, let us sketch the situation.

Recall that when a particle is projected horizontally, we model it as having

zero acceleration (and hence a constant velocity) in the horizontal direction;
zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.

Since the particle is projected horizontally, it has zero initial velocity in the vertical direction. Its acceleration due to gravity is 𝑔=9.8/ms downward. We substitute these two values, along with the distance of 42 m the particle must travel to reach the ground, into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡. Notice that the direction of acceleration is the same as the direction of motion, so the signs on 𝑎 and 𝑠 should be the same. Hence, we have 42=0×𝑡+12×9.8×𝑡𝑡=424.9.

The time taken is, therefore, 𝑡=√8.571…seconds, which is 2.9 seconds to one decimal place.

We can also use the equations of motion to find the horizontal distance traveled by a horizontally projected particle.

An arrow is fired horizontally from a bow at a target at a speed of 74 m/s. The arrow hits the target at a point 15 cm below the point from which it left the bow. Modeling the arrow as a projectile moving freely under gravity 𝑔=9.8/ms in a vertical plane perpendicular to the plane of the target, find the horizontal distance between the bow and the target. Give your answer to two decimal places.

Answer

First, let us sketch the situation.

Recall that when a particle is projected horizontally, we model it as having

zero acceleration (and hence a constant velocity) in the horizontal direction;
zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.

We first analyze the vertical motion of the arrow to find the total flight time of the arrow. This will allow us to calculate the horizontal distance traveled. Thus, we substitute the initial vertical velocity of 0, the acceleration due to gravity of 𝑔=9.8/ms, and the vertical distance of 0.15 m into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡. Notice that the direction of acceleration is the same as the direction of motion, so the signs on 𝑎 and 𝑠 should be the same. Hence, we have 0.15=0×𝑡+12×9.8×𝑡𝑡=0.154.9𝑡=0.1749….

We now substitute this time into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡 using the initial horizontal velocity of 𝑢=74/ms and the horizontal acceleration of 0: 𝑠=74×0.1749…+12×0×(0.1749…)𝑠=12.9473…, giving us a horizontal distance of 12.95 m to 2 decimal places.

Notice that if we are given the time of a projectile’s flight, we can calculate both its horizontal and vertical displacements on landing.

A rock was thrown horizontally from the top of a tower at 20.8 m/s. It flew for 2.4 seconds before hitting the ground. Calculate the distance 𝑠 between the base of the tower and the point where the rock landed, and find the height ℎ of the tower. Take 𝑔=9.8/ms.

Answer

First, let us sketch the situation.

Recall that when a particle is projected horizontally, we model it as having

zero acceleration (and hence a constant velocity) in the horizontal direction;
zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.

Let us calculate the horizontal distance 𝑠 first. We know that the rock’s initial horizontal velocity is 20.8 m/s, its horizontal acceleration is zero, and its time of flight is 2.4 seconds. We substitute these values into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 𝑠=20.8×2.4+12×0×2.4=49.92.

Thus, the distance 𝑠 between the base of the tower and the point where the rock landed is 49.92 metres.

Next, we want to calculate the vertical height of the tower. We know that the rock’s initial vertical velocity is 0, its vertical acceleration is 9.8 m/s2, and its time of flight is 2.4 seconds. We can take the direction of vertical motion as being the same as the direction of vertical acceleration (namely, down). This means that our displacement and acceleration have the same sign. Then, we substitute these values into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 𝑠=0×2.4+12×9.8×2.4=4.9×5.76=28.224.

The height ℎ of the tower is, therefore, 28.224 metres.

On the other hand, if we are given the horizontal and vertical distances of a projectile’s travel, we can calculate its initial velocity.

A ball is thrown horizontally from the top of a tower of length 150 m. It lands on the ground at a horizontal distance of 100 m from the base of the tower. Find the initial velocity at which the ball is thrown, taking the acceleration due to gravity 𝑔=9.8/ms. Give your answer to two decimal places.

Answer

First, let us sketch the situation.

Recall that when a particle is projected horizontally, we model it as having

zero acceleration (and hence a constant velocity) in the horizontal direction;
zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.

We first analyze the vertical motion of the ball. This will allow us to calculate the total flight time. Using this time, we will be able to calculate the initial (horizontal) velocity of the ball.

The ball has an initial vertical velocity of 0, has a vertical acceleration of 9.8 m/s2 due to gravity, and travels 150 m down. We substitute these values into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 150=0×𝑡+12×9.8×𝑡𝑡=1504.9𝑡=5.5328….

We now analyze the horizontal motion of the ball using the fact that the total flight time is 𝑡=5.5328…. The ball has a horizontal acceleration of 0 and travels 100 m horizontally. We substitute these values into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 100=𝑢×5.5328…+12×0×(5.5328…)𝑢=1005.5328…=18.074….

The initial velocity with which the ball was thrown is, therefore, 18.07 m/s to 2 decimal places.

We can also use our understanding of horizontal motion to solve more complicated problems involving friction.

A brick of mass 3 kg is projected along a rough plane at 10 m/s. After traveling for 8 m, the brick leaves the plane and falls 2.5 m to the ground. The total time of motion from the moment of projection to landing on the ground is 2 s, and the acceleration of gravity is 𝑔=9.8/ms.

Find, to one decimal place, the total time that the brick is in contact with the rough plane.
Find, to one decimal place, the coefficient of friction between the brick and the plane.
Find, to one decimal place, the horizontal distance from the point of projection to the point where the brick lands.

Answer

Let us begin by drawing a sketch.

Recall that when a particle is projected horizontally, we model it as having

zero acceleration (and hence a constant velocity) in the horizontal direction;
zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.

Part 1

For part 1, we want to calculate the total time for which the brick is in contact with the rough plane. To do this, we will calculate the time it takes for the brick to fall to the ground after leaving the plane. Then, we will subtract this from the total time of motion, which is given as 2 seconds.

When the brick leaves the plane, it is traveling horizontally. This means that it has a vertical velocity of 0 at this point. After leaving the plane, the brick moves freely under gravity, so its acceleration is 9.8 m/s2 downward. It travels 2.5 m down to the ground. We substitute these values for 𝑢, 𝑎, and 𝑠 into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 2.5=0×𝑡+12×9.8×𝑡𝑡=2.54.9𝑡=0.7142….

Therefore, the total time for which the brick is in contact with the plane is 2−0.7142…=1.2857…, giving us an answer of 1.3 seconds to 1 decimal place for part 1.

Moving forward with the calculation, we will take 𝑡=1.286 to ensure we do not get compounded rounding errors in our final answers to parts 2 and 3.

Part 2

For part 2 we need to find the coefficient of friction between the brick and the plane. Recall that for a body in motion on a rough surface, there is a force of friction acting in the opposite direction to the direction of motion. The magnitude of the friction force is given by the equation 𝐹=𝜇𝑁max, where 𝜇 is the coefficient of friction between the body and the surface and 𝑁 is the normal reaction force. Recall that the normal reaction 𝑁 is equal in magnitude to the gravitational force exerted by the body upon the surface, 𝑀𝑔.

By Newton’s second law, we have 𝐹=𝑀𝑎max, where 𝑀=3kg is the mass of the brick and 𝑎 is the acceleration of the brick (which is in the opposite direction to its motion due to friction). On the other hand, by Newton’s third law, we have 𝑁=𝑀𝑔, where 𝑀=3kg is the mass of the brick and 𝑔 is its acceleration due to gravity. Putting all of this information together, we have first the equation of friction 𝐹=𝜇𝑁,max into which we can substitute 𝐹=𝑀𝑎max and 𝑁=𝑀𝑔, giving 𝑀𝑎=𝜇𝑀𝑔.

Dividing through by the mass 𝑀 and rearranging, we have 𝜇=𝑎𝑔.

Since we are taking 𝑔=9.8 in order to work out 𝜇, we just need to calculate the acceleration of the brick due to friction. We know that the initial horizontal velocity of the brick is 10 m/s and that it travels the 8 m to the edge of the plane in 1.286 seconds. Therefore, we can substitute these values into the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡: 8=10×1.286+12×𝑎×1.286𝑎=2(8−10×1.286)1.286=−5.877…/.ms

Notice that this acceleration is negative, as we should expect, since it represents the deceleration of the brick due to friction. Taking “up” to be the positive direction so that the acceleration due to gravity is negative, we have the coefficient of friction given by 𝜇=𝑎𝑔=−5.877…−9.8=0.59….

Thus, we have 𝜇=0.6 to one decimal place as our answer for part 2. Observe that 𝜇 has no units.

Part 3

Finally, for part 3, let us calculate the total horizontal distance traveled by the brick. We know that the brick first travels for 8 metres along the plane. We now need to calculate its horizontal flight distance as it falls from the plane to the ground. In order to do this, we need to work out its horizontal velocity 𝑉 as it leaves the plane.

Recall that we are rounding quantities to 3 decimal places in the following calculations since that is an acceptable degree of accuracy given that our final answer is only required to be accurate to 1 decimal place. We know that the brick’s initial horizontal velocity is 10 m/s, that it travels along the plane for 1.286 seconds, and that its acceleration during this time is -5.877 m/s2. We substitute these values into the equation of motion 𝑣=𝑢+𝑎𝑡: 𝑉=10+(−5.877)×1.286=10−7.557…=2.442….

Thus, the horizontal velocity of the brick as it leaves the plane is 2.442 m/s to 3 decimal places. Once the brick has left the plane, there are no horizontal forces acting upon it; it is moving freely under gravity; that is to say, its horizontal acceleration is 0. We once again use the equation of motion 𝑠=𝑢𝑡+12𝑎𝑡, with 𝑎=0, 𝑢=𝑉 (the brick’s horizontal velocity as it leaves the plane) and 𝑡=0.714 (the time the brick takes to reach the ground, which we calculated in part 1): 𝑠=𝑉𝑡+0=2.442×0.714=1.743….

Therefore, the brick travels 1.743… metres horizontally from the edge of the plane. This implies that the total horizontal distance from the point of projection to the point where the brick lands is 8+1.743…=9.7metres to 1 decimal place.

Let us finish by recapping a few important concepts from this explainer.

When a particle is projected horizontally, we model it as having zero acceleration (and hence a constant velocity) in the horizontal direction.
In this situation, the particle has zero initial velocity and a constant acceleration of 𝑔 downward in the vertical direction.
We can use these facts, in combination with the equations of motion, to solve problems about horizontal projectile motion.