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