Question

A body projected vertically with a velocity “u” from the ground . it’s velocity at ¾ th of maximum height is

Answer 1

Concept:

• Vertical motion
• One dimensional motion

Given:

• Initial velocity = u

Find:

• The velocity at 3/4th of the maximum height

Solution:

The maximum height is achieved when the final velocity v = 0

v = u +at

0 = u-gt

t =u/g

Maximum height = s = ut -1/2gt^2

s = u²/g - 1/2u²/g = 1/2u²/g

3/4th the maximum height h = 3/8u²/g

To find the velocity at that point,

v² = u²-2gh

v² = u²-2g(3/8)u²/g

v² = u²- 3/4u²

∨²= 1/4u²

v = u/2

The velocity is u/2 at 3/4th the maximum height.

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