A ball is gently dropped form a height of 40m. If its velocity increases uniformly at the rate of 10ms-2, with what velocity will it strike the ground? After what time will it strike the ground
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we have
a = 10m/s^2
u = 0
v = v
S = 40m
so according to formula
v^2 - u^2 = 2aS
v^2 - 0^2 = 2*10*40
v^2 = 800
v = √(800)
v = 20 √(2) m/s
now according to the formula
S = ut + 1/2 at^2
40 = 0(t) + (10)*t^2
40/10 = t^2
4 = t^2
t = 2 seconds.
hope this helps.
mark as brainliest.
a = 10m/s^2
u = 0
v = v
S = 40m
so according to formula
v^2 - u^2 = 2aS
v^2 - 0^2 = 2*10*40
v^2 = 800
v = √(800)
v = 20 √(2) m/s
now according to the formula
S = ut + 1/2 at^2
40 = 0(t) + (10)*t^2
40/10 = t^2
4 = t^2
t = 2 seconds.
hope this helps.
mark as brainliest.
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