What is the maximum height attained by a body which is thrown upward with an initial velocity u?

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CONCEPT:

• Equation of Kinematics: These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

v = u + at

\(s =ut+\frac{1}{2}{at^{2}}\)

 v = u + 2as

where, u = Initial velocity, v = Final velocity, g = Acceleration due to gravity, t = time, and h = height/Distance covered

where u = Initial velocity of the particle at time t = 0 sec

v = Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

EXPLANATION:

Given that;

Initial velocity = u

Acceleration of the body (a) = - g

At maximum height, final velocity () = 0

• At maximum height, the velocity of the body will be zero.
• As the acceleration due to gravity is always in a downward direction and the body is thrown in an upward direction. So the acceleration of the body will be negative.

Use the equation;

⇒ v2 = u2 + 2aS

⇒ 0 = u2 + 2 (- g) × S

⇒ u2 = 2 g S

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A body is thrown vertically upwards with an initial velocity u. Write the expression for the maximum height attained by the body.

A body is thrown vertically upwards with an initial velocity u. Write the expression for the maximum height attained by the body.

v2 = u2 - 2gh

We - get ,

`h_max = u^2/(2g)`

Concept: Newton's Third Law of Motion

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A body is thrown vertically upward with an initial velocity u. Write an expression for the maximum height attained by the body.

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