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.

Asked by Aarthi Vishwanathan | 8th May, 2012, 10:03: PM

Expert Answer:

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

Answered by  | 9th May, 2012, 09:26: AM

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