An outfielder throws a baseball from the outfield to the third baseman. In flight, the ball reaches a height of 25 ft above the ground. The outfielder releases the ball from a height of 6 ft above the ground with a speed of 124 ft/sec. The ball is caught 6 ft above the ground by the third baseman. Determine the horizontal distance between the point where the ball is released and where it is caught

The ball actually rises H = 19 feet, and falls an equal distance. The travel time of the baseball is twice the time it takes to rise or fall 19 feet. That total time is 2*sqrt[H/(2g)]= 1.086 seconds

The vertical velocity component when thrown is (0.543s)*g = 17.5 ft/s

The horizontal velocity component is (and remains) sqrt[124^2 - 17.5^2] = 122.8 ft/s

That should be multiplied by the flight time to get the distance of the throw.

Thanks

To determine the horizontal distance between the point where the ball is released and where it is caught, we can use the equation of motion and kinematic equations.

First, let's consider the vertical motion of the ball. We know that the initial vertical position (y0) of the ball is 6 ft above the ground, the maximum height (ymax) is 25 ft above the ground, and the final vertical position (y) when caught is 6 ft above the ground. We need to find the time it takes for the ball to reach its maximum height and the time it takes for the ball to fall back to the height at which it is caught.

The equation for the vertical position of an object under constant acceleration is given by:
y = y0 + v0t + (1/2)at^2

Since the ball is thrown straight up, the acceleration (a) due to gravity is -32 ft/s^2 (taking downward as the positive direction), the initial vertical velocity (v0) is 0 ft/s (as the ball is released), and the time (t) to reach its maximum height can be determined. So, we get:
ymax = y0 + v0t - (1/2)gt^2
25 = 6 + 0 - (1/2)(32)t^2
19 = 16t^2
t^2 = 19/16
t ≈ 1.16 s

Now, let's find the total time of flight. Since the ball is thrown up and then falls back down, the total time of flight is twice the time it takes to reach the maximum height:
T = 2t
T = 2(1.16)
T ≈ 2.32 s

Next, we can find the horizontal distance (x) covered by the ball during this time. The equation for horizontal motion without any vertical acceleration is given by:
x = v0xT

Since the ball is thrown horizontally, the initial horizontal velocity (v0x) is equal to the magnitude of the initial velocity. So, we have:
x = v0xT
x = (124 ft/s)(2.32 s)
x ≈ 287.68 ft

Therefore, the horizontal distance between the point where the ball is released and where it is caught is approximately 287.68 feet.