Question

From a point 96 ft above the ground, a stone is thrown in such a way that it is at the same point: 3 sec after it was thrown, as it was 2 sec after it was thrown. How long does it take the stone to reach the ground?

Answer 1

Given,

Initial height of stone from ground = 96ft. = 29.26m

Let, the initial velocity of ball be u.

Since the stone is thrown verticlly above, After 2 sec. ,

Velocity of stone =( )

It took (3 - 2) = 1 sec. for the stone to reach the same point as it was before.

So, it has reached its maximum height in 0.5 sec.

At 2 sec. the velocity of stone was ( )

After 2.5 sec. stone has reached maximum height. So,

From kinematics equation,

v = u + at, where v is final vel. i.e. 0 and u is initail velocity (this time we have taken the velocity at 2 sec. as our initaial vel.) and t is time taken and t=0.5

0 = (u - 2*g) + (-g * 0.5)

we get, u = 2.5g

Time taken to reach maximum height is 2 + 0.5 = 2.5 sec.

Time taken to reach from the height to ground,

Using kinematics second equation,

putting s = 29.26m, u = 2.5g and   and a = g, we get,

t = 0.98 sec.

total time taken to reach the ground = 2.5+0.98 = 3.48 sec