Answer:
Explanation:
To solve this problem, we can use the principle of conservation of energy. The initial potential energy of the ball when it is dropped is equal to the final potential energy of the ball when it rebounds.
The initial potential energy of the ball is given by:
PE_initial = m * g * h_initial
where m is the mass of the ball, g is the acceleration due to gravity, and h_initial is the initial height.
The final potential energy of the ball is given by:
PE_final = m * g * h_final
where h_final is the height to which the ball rebounds.
Since the mass of the ball cancels out in both equations, we can equate the two expressions for potential energy:
PE_initial = PE_final
m * g * h_initial = m * g * h_final
h_initial = h_final
Now, let's substitute the given values. The initial height (h_initial) is 1.15 m, and the rebound height (h_final) is 0.815 m.
1.15 m = 0.815 m
Next, we can use the kinematic equation to calculate the velocity at which the ball hits the ground.
The equation is:
v^2 = u^2 + 2 * a * s
where v is the final velocity, u is the initial velocity (which is zero because the ball is dropped), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the negative of the initial height).
v^2 = 0 + 2 * (-9.8 m/s^2) * (-1.15 m)
v^2 = 2 * 9.8 m^2/s^2 * 1.15 m
v^2 = 22.54 m^2/s^2
Taking the square root of both sides, we get:
v = √(22.54 m^2/s^2)
v ≈ 4.75 m/s
Therefore, the velocity at which the tennis ball hits the ground is approximately 4.75 m/s.