A home run is hit in such a way that the baseball just clears a wall 25.0 m high, located a distance 125 m from home plate. The ball is hit at an angle of 38.0° above the horizontal, and air resistance is negligible. (Assume the ball is hit at a height of 1.0 m above the ground.)
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To determine the initial velocity of the baseball when it was hit, we can use the following kinematic equation:
h = (v_initial * sin(theta_initial) * t) - (1/2 * g * t^2)
Where:
- h is the vertical height the baseball reaches (25.0 m)
- v_initial is the initial velocity of the baseball
- theta_initial is the launch angle (38.0°)
- t is the time it takes for the baseball to reach the peak and fall back down
- g is the acceleration due to gravity (9.8 m/s^2)
Since the baseball starts and ends at the same height (1.0 m), the equation becomes:
24.0 m = (v_initial * sin(38.0°) * t) - (1/2 * g * t^2)
We also know that the horizontal distance the ball travels is 125 m. The equation for horizontal motion is:
x = v_initial * cos(theta_initial) * t
Where:
- x is the horizontal distance the baseball traveled (125 m)
By rearranging the equation, we can solve for t:
t = x / (v_initial * cos(theta_initial))
Now we can combine the two equations to solve for v_initial:
24.0 m = (v_initial * sin(38.0°) * (x / (v_initial * cos(38.0°)))) - (1/2 * g * (x / (v_initial * cos(38.0°)))^2)
Simplifying the equation gives:
24.0 m = (x * tan(38.0°)) - (1/2 * g * (x / v_initial)^2)
Rearranging the terms will isolate v_initial:
(v_initial^2 * sin^2(38.0°)) + 2 * g * x * tan(38.0°) = x^2 * g
v_initial^2 = (x^2 * g - 2 * g * x * tan(38.0°)) / sin^2(38.0°)
Taking the square root of both sides will give us the magnitude of the initial velocity:
v_initial = sqrt((x^2 * g - 2 * g * x * tan(38.0°)) / sin^2(38.0°))
Now we can substitute the given values into the equation and solve for v_initial:
v_initial = sqrt((125^2 * 9.8 - 2 * 9.8 * 125 * tan(38.0°)) / sin^2(38.0°))
Calculating this expression results in the value of v_initial = 35.5 m/s.