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A stone of mass m is dropped from rest at a height of 2point88 m From what height would a stone of mass m divided by 2 have to be dropped to have the same momentum upon striking the ground?
Wiki User
∙ 13y ago
x=1/2at2 and
a=9.81m/s2 so
2.88m = 4.905t2
t = 0.77 s
v=at so
v= 7.5 m/s
momentum = mv = (m)kg* 7.5m/s = 7.5m
Now with a mass of m/2 having a momentum of 7.5m we have to double the velocity.
7.5m = 7.5(m/2)*2 = 15m
To get a velocity of 15m/s
15m/s = at so t= 1.53 s
distance is x=1/2at2
x=1/2at2 x= 11.48m
Wiki User
∙ 13y ago
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