Question

Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the horizontal. The ball is struck 3 feet above home plate. You watch as the ball goes over the outfield wall 420 feet away and lands in the bleachers. After you congratulate Ted on his hit he tells you, 'You think that was something, if there was no air resistance I could have hit that ball clear out of the stadium!' Assuming Ted is correct, what is the maximum height of the stadium at its back wall x = 565 feet from home plate, such that the ball would just pass over it? You may need: 9.8 m/s2 = 32.2 ft/s2 1 mile = 5280 ft

Answer 1

Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°)  * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth  the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - $\frac{32.2 ft/s^{2} *3.92 s^{2} }{2}$

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32