A ball is gentle dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground?

Asked by Aryan Suresh | 25th Jun, 2013, 09:13: PM

Expert Answer:

s = 20m
u = 0m/s
a = 10m/s^2
 
Using third law of motion
v^2 = u^2+2as
v^2 = 0^2 + 2(10)(20)
v^2 = 400
v = 20m/s
 
Also, using first law of motion
v = u + at
20 = 0+10t
t = 2 sec

Answered by  | 26th Jun, 2013, 04:16: AM

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