Map sapling.learming this question was wrirten by John Carlson at University of
ID: 1502108 • Letter: M
Question
Map sapling.learming this question was wrirten by John Carlson at University of Colorado, Denver A solid uniform sphere rolls without slipping up a hill beginning at its base with 26.0m/s. The hill rises and then levels off before ending in a vertical cliff. It rises a total of 27.0m. The ball goes up the hill and then launches horizontally off cliff and landing at the original height at the hill's base. The ball's radius is 5.00cm. How fast is the ball traveling at the top of the hil? Number 17.25 m/ s How far from the base does it land? Number 40.47 How fast is it traveling when it hits the ground? Number 28.76 m/ s What is the difference in angular speed between angular speed at the beginning of the problem at the base of the hill and the speed at the bottom of the fall? Number 174.656 rad/s Previous Check Answer Next ExitExplanation / Answer
The initial kinetic energy of the rolling ball is
0.5*I*w^2 + 0.5*m*v^2. If you put in w = v/r and I = 2/5 * m*r^2, you get total initial kinetic energy of 0.7*m*v0^2. Going up the hill, the remaining kinetic energy is 0.7*m*v0^2 - m*g*h, so the speed of the ball at the top is 0.7*m*v^2 = 0.7*m*v0^2 - m*g*h,
v = [v0^2 - g*h/0.7] = a
This is the horizontal speed of the ball.
The time it takes for the ball to reach the ground is given by h = 0.5*g*t^2, or t = [2*h/g].
The horizontal distance traveled in that time is then x = v*t = [v0^2 - g*h/0.7] * [2*h/g].
The velocity at landing is the vector sum of the horizontal velocity v (from above) and the vertical velocity = g*t = [2*h*g] = b
The vector sum is the square root of the sum of the squares of the two velocities: [a^2 + b^2] (a, b are found above)