A tennis ball, starting from rest, rolls down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x.

3.61metres

Step-by-step explanation:

The first thing to do is that as the ball rolls down the hill, its initial potential echanged into rotational and linear

Assuming the height is 3.2 meters

kinetic energy.

Initial Potential Energy = m * 9.8 * 3.20 = m * 31.36

For a thin walled spherical shell, the moment of inertia is ⅔ * m * r^2

Rotational KE = ½ * I * ω^2 = ⅓ * m * r^2 * ω^2

ω = v/r, ω^2 = v^2/r^2

Rotational KE = ⅓ * m * v^2

Linear KE = ½ * m * v^2

Total KE = 5/6 * m * v^2

5/6 * m * v^2 = m * 31.36

v^2 = 37.632

v = √37.362

The ball’s velocity at the bottom of the hill is approximately 6.134 m/s. To determine he range, use the following equation.

Let R be equal to the range

R = v^2/g * sin 2 θ

R = 37.632/9.8 * sin 70 = 3.84 * sin 70

Which is equal to 3.61 meters.

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

answer: 12x^2+6x

step-by-step explanation:

6x does not equal 24x. hope it works!