h = 0.5g*t^2 = 7.5
4.9t^2 = 7.5
t^2 = 1.53
t = 1.24 s. = Fall time.
d = Vx*t = 15m/s * 1.24s = 18.6 m.
A package is dropped from a helicopter flying at a height of 7.5 m while traveling 15 m/s. How far is the package travel horizontally before it hits the ground below?
2 answers
h = 0.5g*t^2 = 7.5
4.9t^2 = 7.5
t^2 = 1.53
t = 1.24 s. = Fall time.
d = Vx*t = 15m/s * 1.24s = 18.6 m.
1 0
4.9t^2 = 7.5
t^2 = 1.53
t = 1.24 s. = Fall time.
d = Vx*t = 15m/s * 1.24s = 18.6 m.
1 0