acceleration = 9.81m/s^2 down
initial height= Hi = 50 m
initial velocity = Vi = 0
h = Hi + Vi t + (1/2) a t^2
0 = 50 + 0 t - 4.9 t^2
sp
t = sqrt (50/4.9)
t = 3.19 s
so how fast?
v= Vi + a t
v = 0 - 9.81 * 3.19 = -31.3 meters / second
A body mass 2kg falls freely through a height of 50m and comes to rest having penetration 5.0 of sand Cal
The velocity with which the body hits the sand
1 answer