**Question:**

**A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity **

**of half the first. The vertical gap between first and second ball is +15m at t = 2s. The gap is found to remain constant. Calculate the velocity with which the balls **

**were thrown and the exact time interval between their throw.**

**Solution:**

Let the speed of ball 1 = u1 = 2u m/s

Then the speed of ball 2 = u2 = u m/s

The height covered by ball 1 before coming to rest = h1

The height covered by ball 2 before coming to rest = h2

We know that,

v2 = u2 + 2gh

v2 = 2gh

h = v2/2g

Therefore, h1 = 4u2/2g

h2 = u2/2g

From question, h1 –h2 = 15 m

Therefore,

4u2/2g – u2/2g = 15

u = 10 m/s

h1 = 20 m

h2 = 5 m

And time taken is,

t1 = 2 sec

t2 = 1 sec