Question
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
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.