From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25m/s. Find when and where the two balls meet.

# From the top of a tower 100 m in height a ball is dropped and at the same

for the ball dropped downwards

s=ut+1/2gt^{2}

s=1/2gt^{2}

s=4.9*t ^{2}
For the ball thrown upwards
= 100-s=ut-1/2gt^{2}
100-s=25*t-4.9t

^{2}

therefore

100-s=25t-49.t

^{2}

Now add both the equations

100 = 25 t

So t = 4 seconds

S = 4.9 *16 m