physics help?

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 = 0.66 m from the end of the track after falling a vertical distance h2 = 1.20 m from the end of the track.




(a) At what height above the ground does the ball start to move?
m
(b) What is the speed of the ball when it leaves the track?
m/s
(c) What is the speed of the ball when it hits the ground?
m/s

Answer 1

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For horizontal projectile,

Time of flight =T = sq rt (2h/g)

Horizontal range =R =u*sq rt (2h/g)

u=R*sq rt g/2h

The ball leaves the track horizontally, striking the ground a distance x = 0.66 m from the end of the track after falling a vertical distance h2 = 1.20 m from the end of the track

R=distance x =0.66 m

h =vertcal distance h2 =1.20 m

u =R*sq rt g/2h

u =0.66* sq rt [9.8/(2*1.2) ]

u =0.66*2.0207

u = 1.333 m/s

the ball leaves the track with speed 1.333 m/s

vertical distance the ball moved on the track=h1

As the ball starts from rest, h1=(velocity at end of track)^2/2g

h1 =1.333*1.333 /2*9.8=0.09075 m

Total height =h1+h2=0.09075 +1.2=1.29075 m

(a)The ball starts to move at height 1.29075 m above the ground
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(b) The speed of the ball is 1.333 m/swhen it leaves the track
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(c)The speed of the ball when it hits the ground =v = sq rt [u^2+2gh]

The speed of the ball when it hits the ground =v = sq rt 25.297

The speed of the ball when it hits the ground =v = 5.027 m/s

(c)The speed of the ball when it hits the ground is 5.027 m/s
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Answer 2

a) Potential energy at the top (height h) equal to kinetic energy at the time the ball leaves the slide
Pe=Ke and
mgh=0.5mV^2 where
m - mass of the ball
g - acceleration due to gravity

we have
h= 0.5mV^2 / mg= V^2 / (2g)
Now we have to find V. V is the horizontal velocity Vh.

since Vh=1.3
h1= (1.3)^2 / (2 x 9.81)=0.086m

b) Vh = S/t
t=sqrt(2 h2/g)
Vh= S/sqrt(2 h2/g)
Vh= S sqrt(g/ (2 h2) )where
h2 - is the height of the slide's release point above the ground

Vh= 0.66 sqrt(9.81/(2 x 1.2))
Vh=1.3 m/s

c) V=sqrt(Vv^2 + Vh^2)
Vertical speed
Vv=gt
Vv= g sqrt(2 h2/g)
Vv= sqrt(2 g h2)

Vv=sqrt(2 x 9.81 x 1.2)
Vv=4.9m/s
finally
V= sqrt(4.9 ^2 + 1.3^2)=5.0m/s