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A ball is thrown horizontally from a cliff at a speed of 15 ms and strikes the ground 45 meters from the base of the cliff How high was the cliff?
Wiki User
∙ 14y ago
KE before collision =1/2 m v2 = 1/2 m (15)2= 1/2 m 225 (Velocity)2 after collision v2=u2 +2as Final velocity is 0 when object is at maximum height u is the initial velocity just after the rebound a is due to gravity = -9.8ms-2 s is the displacement = 5m 0 = u2 + 2(-9.8)5 u2=98 (ms-1)2 KE after rebound = 1/2 m 98 since mass does not change Fraction of KE remaining = 98/225 Fraction lost 1 - fraction remaining 1- (98/225) = 127/225 Not: if g is taken to be 10ms-2, then fraction is 125/225 = 4/9
Wiki User
∙ 15y ago
Wiki User
∙ 13y agoanswer= 1.06kg*m/s
max height = zero
because its not moving, so m*v=0
then,
v^2=Vinitial^2+2ad
Vfinal^2=Vinitial^2+2ad
0=225-19.6*d
Vinitial=15
Vfinal= 0
a= -9.8
d=225/19.6=11.47m
halfway = 5.37m
soooo
V^2=15^2+2*-9.8*5.73 (to know half)
v^2=122.692
v= sqrt (112.692)
= 10.6m/s
p=mv
p=(0.1)(10.6)=1.06kg*m/s
By: lgmena
OR
V = (15) √0.5
= 15/1.414
= 10.6 m/s
P= mv
= 1.06 kg m/s
Wiki User
∙ 10y agoinitial KE= final GPE
1/2 m v^2 = mgh
V= sqrt(2gh)
V=sqrt(2 x 9.80m/s^2 x 5m)
V=9.90 m/s upwards
Wiki User
∙ 7y agoA ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff (rounded to the nearest meter)?
Wiki User
∙ 14y agoAnswer: 44 meters
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