A tennis ball is shot vertically upward from the surface of an atmosphere-free p

Question

A tennis ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. A) How long does it take the ball to reach its maximum height? B) How high does the ball rise? C) What is the magnitude of the acceleration due to gravity on the surface of this planet? D) Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive. E) How long has the ball been in the air when it returns to its original position?

Explanation / Answer

Okay, so in one second, the ball has slowed down from 20 m/s to 15 m/s What does this tell us? That the acceleration of the gravity is -5 m/s/s Because of the simple values they give us, it is easy to see that in another 3 seconds, the ball will be stationary, and start to fall. So we know the time that it takes to reach its maximum height, 4 seconds. using the equation of motion x = ut + 1/2at^2 where u = initial velocity = 20m/s t = time of flight = 4 seconds a = acceleration = gravity = -5m/s/s we get this: x = 20 * 4 + 0.5 * -5 * 4^2 x = 80 - 40 x = 40 meters.

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