Asked by Mike
Hoping Henry can help on previous prob.
A ball is thrown upwards from a roof top, 80ft above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time T is H, which is given by H= -16t2 + 64t +80.
V = Vo + g*Tr = 0.
64 + 32Tr = 0,
Tr = 2 s. = Rise time or time to reach max height, H.
H = -16*2^2 + 64*2 + 80 = 144 Ft.
0.5g*Tf^2 = 144.
16Tf^2 = 144,
Tf = 3 s. = fall time.
T = Tr + T = 2 + 3 = 5 s. = Time in flight.
Please elaborate,
How did you come up with 3s for the fall time?
A ball is thrown upwards from a roof top, 80ft above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time T is H, which is given by H= -16t2 + 64t +80.
V = Vo + g*Tr = 0.
64 + 32Tr = 0,
Tr = 2 s. = Rise time or time to reach max height, H.
H = -16*2^2 + 64*2 + 80 = 144 Ft.
0.5g*Tf^2 = 144.
16Tf^2 = 144,
Tf = 3 s. = fall time.
T = Tr + T = 2 + 3 = 5 s. = Time in flight.
Please elaborate,
How did you come up with 3s for the fall time?
Answers
Answered by
Steve
max height reached at t=2
h=0 at t=5
so, 2 seconds rising, 3 seconds falling
he got the 3 by seeing how long it takes to fall 144 ft. s = 1/2 at^2
h=0 at t=5
so, 2 seconds rising, 3 seconds falling
he got the 3 by seeing how long it takes to fall 144 ft. s = 1/2 at^2
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