Question
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
The velocity with which the body hits the sand
Answers
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
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
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