Asked by Tim
A tennis ball, starting from rest, rolls down a hill. At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin walled spherical shell (I=2/3 mr^2 )and determine the range x it travels if it starts at a height of 2.50 m
Answers
Answered by
Damon
You need how far it rolled down the hill to get initial speed
v = w r where w is omega, the angular velocity
kinetic energy = m g h
= (1/2)I w^2 + (1/2) mv^2
= (1/2)(2/3)mr^2v^2/r^2 + (1/2)mv^2
=(mv^2/2)[2/3 + 1] = (5/3) (mv^2/2)
so
g h = (5/3)(v^2/2)
solve for v, initial launch speed
Vi = v sin 35
u = v cos 35
then it is conventional parabolic flight problem.
v = w r where w is omega, the angular velocity
kinetic energy = m g h
= (1/2)I w^2 + (1/2) mv^2
= (1/2)(2/3)mr^2v^2/r^2 + (1/2)mv^2
=(mv^2/2)[2/3 + 1] = (5/3) (mv^2/2)
so
g h = (5/3)(v^2/2)
solve for v, initial launch speed
Vi = v sin 35
u = v cos 35
then it is conventional parabolic flight problem.
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