Question

A particle is projected vertically upwards from point A with an initial speed of 29.4 m/s. During what intervals of time is the particles more than 39.2 m above A?
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Answer 1

. Consider the motion from A to B:

s = +H (final point lies above the initial point), initial velocity = u, final velocity v = 0.

Let the time taken to go from A to be t 1

Using v = u - gt, we get

0 = u - gt¹

⇒ t1 =gu

Using s = ut - (1/2)gt²

H = u( gu ) - 21

g ( g ) 2 = 2gu 2

So the maximum height attained is H =

2g

u

2

Consider the return motion from B to A:

s = -H (final point lies below the initial point)

u = 0 (at point B, velocity is zero)

Let time taken to go from B to A be t 2

. We have t2 = gg2 = g2g²u 2 = gu

proved.

c. Displacement = 0, distance travelled = 2H = 2gu²

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