A ball of mass 522 g starts at rest and slides down a frictionless track?

Question

It leaves the track horizontally, striking the ground a distance x = 0.95 m from the end of the track after falling a vertical distance h2 = 1.03 m from the end of the track.
(a) At what height above the ground does the ball start to move?
(b) What is the speed of the ball when it leaves the track?
(c) What is the speed of the ball when it hits the ground?

Answer 1

Equations of motion
y(t)=1.03-.5*9.81*t^2
x(t)=vi*t

when y=0 and x=0.95, the ball strikes the ground

0=1.03-.5*9.81*t^2
0.95=vi*t

solve for vi

vi=2.07 m/s

a) .5*m*v^2=m*g*h
or
h=.5*v^2/g
h=0.219 m

b) v=vi=2.07m/s

c) Speed at the ground requires we look at horizontal: 2.07 and vertical:
vy=9.81*0.458
vy=4.5 m/s
so the magnitude of velocity is
sqrt(4.5^2+2.07^2)
speed=4.95 m/s

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