Final answer:
The velocity of a tennis ball when it hits the ground is approximately -5.77 m/s. The velocity of the tennis ball when it leaves the ground and begins its rebound is approximately -5.90 m/s. The acceleration provided by the ground during contact is approximately -9.63 m/s².
Step-by-step explanation:
To calculate the velocity of the tennis ball when it hits the ground, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. In this case, the initial velocity is 0 m/s (since it is dropped), the acceleration is -9.8 m/s^2 (due to gravity acting downwards), and the distance traveled is 1.7 m. Plugging these values into the equation, we get v^2 = 0^2 + 2*(-9.8)*1.7 = 33.32. Taking the square root of both sides, we find that the velocity of the tennis ball when it hits the ground is approximately -5.77 m/s.
To find the velocity of the tennis ball when it leaves the ground and begins its rebound, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is -5.77 m/s (the negative sign indicates that it is moving upwards), the acceleration is -9.8 m/s^2 (due to gravity acting downwards), and the time is 0.0135 s. Plugging these values into the equation, we get v = -5.77 + (-9.8)*0.0135 = -5.90 m/s.
When the tennis ball is in contact with the ground, the acceleration provided by the ground can be calculated using the equation a = (v - u) / t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the final velocity is -5.90 m/s, the initial velocity is -5.77 m/s, and the time is 0.0135 s. Plugging these values into the equation, we get a = (-5.90 - (-5.77)) / 0.0135 = -9.63 m/s^2.