College Physics

Bob is about 157.64 feet up the cliff. Since we know the ball travels at 76.0 mph (34.39 m/s) we can use the equation for uniform acceleration: vf^2 = vi^2 + (2*a*d) where vf is the final velocity of the ball, vi is the initial velocity of the ball (zero, since it is being thrown), a is the acceleration of gravity (9.8 m/s^2), and d is the distance the ball has traveled. We can use this equation to determine the y-component of the initial velocity (vy) of the ball, which is the component that helps the ball travel upwards out of the initial five feet of the cliff's edge. vy = sqrt(2*a*d) = sqrt(2*9.8*5) = 14.14 m/s Since the ball travels upwards at a constant speed of 14.14 m/s and then follows a curved path downwards for 0.176 seconds, we can calculate the total distance traveled by the ball: Distance traveled = (vy * time) + 1/2(acceleration of gravity * time^2) Distance traveled = (14.14 m/s * 0.176 s) + 1/2(9.8 m/s^2 * 0.176 s^2)