Question

Bob has just finished climbing a sheer cliff above a level beach, and wants to figure out how high he climbed. All he has to use is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is about 33.3 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after 0.910 seconds the ball is once again level with Bob. Bob can't see well enough to time when the ball hits the ground. Bob's friend then measures that the ball hit the ground 129 m from the base of the cliff. How high above the beach was the ball when it was thrown?

Answer 1

To find the height above the beach that the ball was thrown, we can use the equation for projectile motion. The height is approximately 56.83 meters.

Step-by-step explanation:

To find the height above the beach that the ball was thrown, we can use the equation for projectile motion. The formula for the height is given by h = (v^2 * sin^2θ) / (2g), where h is the height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

In this case, the ball was thrown straight up, so the launch angle is 90 degrees and the sine of 90 degrees is 1. The acceleration due to gravity is approximately 9.8 m/s^2.

Using the value for the initial velocity of 33.3 m/s, we can substitute these values into the formula to find the height. h = (33.3^2 * 1^2) / (2 * 9.8) = 56.83 meters.