Question

A ball is thrown from the top of the tower in vertically upward direction . Velocity at a point h metre below the point of projection is twice of the velocity at a h metre above the point of projection . Find the maximum height reached by the ballbabove the top of tower . ( Neglect air friction)​

Answer 1

Answer:

3h/2

Explanation:

taking the downward as positive

we have,

v1²=u²-2gh

where u is the initial velocity

v2²=2gh

given:

v2=2*v1

2gh=u²-2gh

u²=3gh

h=u²/3g

so,maximum height is

when

H=u²/2g

=3gh/2g

=3h/2

so,the maximum height is 1.5*h.