Asked by Neal parikh
A ball is thrown vertically upward from the edge of a bridge 29.0 m high with an initial speed of 20.0 m/s. The ball falls all the way down and strikes the water below. Determine the time it takes the ball to strike the water.
Answers
Answered by
Henry
V = Vo + g*Tr = 0.
Tr = -Vo/g = -20/-9.8 = 2.04 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2.
h = 29 + 20*2.04 - 4.9*2.04^2 = 49.4 m Above gnd.
0.5g*Tf^2 = 49.4, 4.9Tf^2 = 49.4, Tf^2 = 10.08, Tf = 3.18 s. = Fall time.
T = Tr + Tf = 2.04 + 3.18 = 5.22 s. To strike the water.
Tr = -Vo/g = -20/-9.8 = 2.04 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2.
h = 29 + 20*2.04 - 4.9*2.04^2 = 49.4 m Above gnd.
0.5g*Tf^2 = 49.4, 4.9Tf^2 = 49.4, Tf^2 = 10.08, Tf = 3.18 s. = Fall time.
T = Tr + Tf = 2.04 + 3.18 = 5.22 s. To strike the water.
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