A solid, uniform ball of mass 16.9 kg and radius 0.35 m, rolls without slipping

Question

A solid, uniform ball of mass 16.9 kg and radius 0.35 m, rolls without slipping up a hill of height 16.5 m, as shown in the figure. At the bottom of the hill the ball has a velocity of 26.1 m/s. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff.
1) What is the total initial kinetic energy of the ball at the bottom of the hill? (Use the numbers in the introduction, not the picture.)
2) What is the velocity of the ball just as it leaves the top of the hill?
3) How long is the ball in the air?
4) How far from the foot of the hill does the ball land?
%0 How fast is it moving just before it lands?

Explanation / Answer

The mass of the ball M = 16.9 kg
the radius of the ball R = 0.35 m the heigth of the hill h = 16.5 m the initial velocity of the ball vi = 26.1 m/s (1) The total kinetic energy            Ki = (1/2) I^2 + (1/2)mv^2                = (1/2)(2/5MR^2)(v/R)^2 + (1/2)Mv^2                = (1/5)(Mv^2) + (1/2)Mv^2                = (7/10)Mv^2                = (7/10)(16.9 kg)(26.1 m/s)^2                = 8058.71 J ..................................................................................... ..................................................................................... 2) The final translational velocity of the ball before it hits the
ground is the vector sum of the X and Y components of velocity after the ball leaves the cliff. Then you must find the time it takes for the ball to
fall to the ground from 16.5 m :
from equation of motion ,          y - yo = vot + (1/2)at2              0 -16.5 m = 0 - (1/2)(9.8 m/s2)(t2)
         -16.5 m = -4.9 (t2)
                   t = 1.835 s
From law of cosnervation of energy                 Ki = Uf + Kf         then Kf = Ki - Uf                     = 8058.71 J - Mgh                    = 8058.71 - (16.9)(9.8)(16.5) (7/10)Mv^2 = 5325.98 then the final speed is
                vx = [(5325.98)(10)/(16.9)(7)]1/2                     = 21.218 m/s velocity of the ball at the top of the hill vx = 21.218 m/s ......................................................................................                     = 21.218 m/s velocity of the ball at the top of the hill vx = 21.218 m/s ...................................................................................... 3)     height h = 16.5 m     velocity vix = 21.218 m/s    time t = 2h/g             = 2*16.5/9.8             = 1.835 s ................................................................. 4) horizontal range R = (vix)2h/g                              = (21.218 m/s)(1.835)                              = 38.93 m .................................................................. 5) Y-component of the velocity :
                 vy = [9.8m/s2](1.835 s)                     = 17.98 m/s ................................................................ ................................................................ So the final velocity of the ball is:               V = [(21.218 m/s)2 + (17.98 m/s)2]                  = 27.81 m/s

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