Answer:
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.
Based on the angle with respect to the ground and the height, the range x is 2.029 m.
The range can be found as:
= (v² x Sin (2 x angle)) / g
Use the work-energy theorem to find v as:
= √(6 x g x h) / 5
= √(6 x 9.8 x 1.8) / 5
= 4.6 m/ sec
The range is:
= (4.6² x Sin (2 x 35)) / 9.8
= 2.029 m
Find out more on ranges at https://brainly.com/question/7644636.
