A ball is gently dropped from a height of 50 metres. If it's velocity increases uniformly at the rate of 10 metre/second square,with what velocity will it strike the ground? After what time will it strike the ground?
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Distance, s = 50m
Acceleration, a = 10m/ s^2
Initial Velocity, u = 0
Final Velocity, v = ?
By third equation of motion,
v^2= u^2 + 2as
v^2 = 0 + 2*10*50
v^2 = 1000
v= 10√10 m/s
By second equation of motion,
s = ut + 1/2 a*t^2
50 = 0 + 1/2 10*t^2
10= t^2
t= √10 seconds
So the ball will reach the ground after √10 seconds with velocity 10√10 metres per second.
Acceleration, a = 10m/ s^2
Initial Velocity, u = 0
Final Velocity, v = ?
By third equation of motion,
v^2= u^2 + 2as
v^2 = 0 + 2*10*50
v^2 = 1000
v= 10√10 m/s
By second equation of motion,
s = ut + 1/2 a*t^2
50 = 0 + 1/2 10*t^2
10= t^2
t= √10 seconds
So the ball will reach the ground after √10 seconds with velocity 10√10 metres per second.
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