Question

2. A body projected vertically up crosses points A and B
separated by 28 m with velocities one-third and one-
fourth of the initial velocity, respectively. What is the
maximum height reached by it above the ground?​

Answer 1

Answer:

ANSWER

Let initial velocity of the body = u

Velocity of the body at point A =

3

u

Velocity of the body at point B =

4

u

Apply v

2

−u

2

= 2aS for points A and B:

(

4

u

)

2

−(

3

u

)

2

=2(−g)×28

On solving we get: u

2

= 1152g

Now apply direct formula for maximum height

H

max

=

2g

u

2

H =

2g

1152g

= 576 m