A ball is dropped gently from a height of 40 m. If velocity increases at the rate of 40 m/sec2 , with what velocity will it strike the ground? After, what time will it strike the ground?
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Distance , s = 40m
Acceleration, a = 40m / second^2'
Initial Velocity, u = 0
v^2 = u^2 + 2as
v^2 = 0 + 2*40*40
v^2 = 3200
v= 40 * 2^(1/2) [ I don't know how to put the root symbol ]
s = ut + 1/2 a*t^2
40 = 0 + 20*t^2
2=t^2
t=2^(1/2) seconds.
So, the ball will hit the ground after 1.414 seconds or 2*(1/2) seconds with velocity 56.56 m/s or 40* 2^(1/2) m/s
Acceleration, a = 40m / second^2'
Initial Velocity, u = 0
v^2 = u^2 + 2as
v^2 = 0 + 2*40*40
v^2 = 3200
v= 40 * 2^(1/2) [ I don't know how to put the root symbol ]
s = ut + 1/2 a*t^2
40 = 0 + 20*t^2
2=t^2
t=2^(1/2) seconds.
So, the ball will hit the ground after 1.414 seconds or 2*(1/2) seconds with velocity 56.56 m/s or 40* 2^(1/2) m/s
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