Physics help please!!!?

Question

A ball of mass 522 g starts at rest and slides down a frictionless track, as shown below. It leaves the track horizontally, striking the ground a distance x = 0.71 m from the end of the track after falling a vertical distance h2 = 1.24 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

Let's find out how long the ball was in the air:
1.24=.5*g*t^2
solve for t
0.503 seconds

the speed of the ball when it leaves the track then is
0.71=vi*0.503
solve for vi
vi=1.41 m/s (this is part b)

a) relate vi to the height of release
.5*m*vi^2=m*g*h
h=.5*vi^2/g
h=0.10163 m

c) the horizontal component is vi
the vertical is
vy=g*t
=9.81*0.503
4.93 m/s
so the resultant is
sqrt(4.93^2+1.41^2)
5.13 m/s

j