Question

A baseball, hit 7 feet above the ground, leaves the bat at an angle of 55 ° and is caught by an outfielder 6 feet above the ground and 48 feet from home plate. Neglecting air resistance, what was the initial speed of the ball? Use g = 32 ft/sec2 as the acceleration of gravity. Your answer should be in feet per second.

Answer 1

To find the initial speed of the ball, use the vertical motion formula to determine the time it takes for the ball to reach the outfielder and the vertical component of the initial velocity. Then, use the vertical motion formula to solve for the initial speed.

Step-by-step explanation:

To find the initial speed of the ball, we can use the vertical motion formula. The vertical distance traveled by the ball is the sum of the initial height and the final height, which is 6 feet + 48 feet. The time it takes for the ball to reach the outfielder can be found using the equation d = vt, where d is the horizontal distance and v is the initial speed. The vertical motion can be separated into two components: initial vertical velocity, which is the initial speed multiplied by the sine of the launch angle, and the vertical acceleration, which is the acceleration due to gravity. Using these equations, we can solve for the initial speed of the ball.

First, we need to find the time it takes for the ball to travel 48 feet horizontally. Using the equation d = vt, we get t = d/v. Plugging in the values, we get t = 48 feet / v.

Next, we can find the vertical component of the initial velocity using the equation v_y = v * sin(θ), where θ is the launch angle. Plugging in the values, we get v_y = v * sin(55°).

Using the vertical motion formula y = y_0 + v_y * t - (1/2) * g * t^2, where y is the final height, y_0 is the initial height, g is the acceleration due to gravity, and t is the time, we can plug in the values and solve for v.

After solving the equation, the initial speed of the ball is approximately 73.13 ft/s.