Question

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor

a) 7.35 m/s
b) 8.20 m/s
c) 9.10 m/s
d) 10.05 m/s

Answer 1

1. Final answer:
2. To find the velocity of the steel ball just before it strikes the floor, we can use the principle of conservation of energy. The potential energy of the ball when it is at a height of 1.50 m is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
4. Step-by-step explanation:
5. To find the velocity of the steel ball just before it strikes the floor, we can use the principle of conservation of energy. The potential energy of the ball when it is at a height of 1.50 m is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. The gravitational potential energy is then converted to kinetic energy when the ball is about to strike the floor. Therefore, we can equate the potential energy at 1.50 m to the kinetic energy just before impact:
7. mgh = 0.5mv^2
9. Plugging in the given values of h = 1.50 m and h' = 1.45 m, we can solve for the initial velocity:
11. 0.075 x 9.8 x 1.50 = 0.5 x 0.075 x v^2
13. Using a calculator, we get v ≈ 7.35 m/s.