Asked by Lea
A man on a 2m high ladder throws a ball in the air with an initial velocity of 22m/s. If the ball is not caught by the man and instead hits the ground, how long was it in the air and how high did it go?
Answers
Answered by
Henry
Tr = (V-Vo)/g = (0-22)/-9.8 = 2.24 s. =
Rise time.
hmax = ho + Vo*t + 0.5g*t^2.
hmax = 2 + 22*2.24 - 4.9*(2.24)^2 = 26.7m. Above gnd.
hmax = Vo*t + 0.5g*t^2 = 26.26.7m.
0 + 4.9t^2 = 26.7
4.9t^2 - 26.7 = 0.
Use Quadratic Formula:
Tf = 2.33 s. = Fall time.
Tr + Tf = 2.24 + 2.33 = 4.57 s. = Time
in air.
Rise time.
hmax = ho + Vo*t + 0.5g*t^2.
hmax = 2 + 22*2.24 - 4.9*(2.24)^2 = 26.7m. Above gnd.
hmax = Vo*t + 0.5g*t^2 = 26.26.7m.
0 + 4.9t^2 = 26.7
4.9t^2 - 26.7 = 0.
Use Quadratic Formula:
Tf = 2.33 s. = Fall time.
Tr + Tf = 2.24 + 2.33 = 4.57 s. = Time
in air.
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