Question

A body is projected vertically upwards with a velocity of
10 m/s. It reaches the maximum height h in time t. In time
t/2, the height covered is:
(a) h/2 (b) (2/5)h (c) (3/4)h (d) (5/8)h​

Answer 1

Explanation:

max height=u²/2g

=5m

t=u/g

1s

s=10(0.5)-5(0.5)²

=5-5/4

(3/4)5

=(3/4)h

Answer 2

So we have to consider only vertical motion. Here,

u = 10 m s^(-1)

a = - g = - 10 m s^(-2)

At a time t, by the second kinematic equation, we have,

h = 10 t - 5 t²

5 t² - 10 t + h = 0

But, by third kinematic equation, h = u² / 2g = 5 m, since v = 0. So,

5 t² - 10 t + 5 = 0

t² - 2 t + 1 = 0

(t - 1)² = 0

This implies that t = 1 s.

So in time 1 / 2, let the body cover the height h'. So,

h' = 10(1 / 2) - 5(1 / 2)²

h' = 5 - (5 / 4)

h’ = 5 (1 - (1 / 4))

h' = 3 h / 4 [because h = 5 m]

Hence the answer is (c).