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?

Question

Shaun Camacho

Physics

7 January, 12:42

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?

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Alvaro Mckay · Answer 1

The equations that we will be using are: y = 0 + v*sin (35) * t - 1/2*g*t^2 andx = v*cos (35) * t

[we have x = 565, v = 176 ft/s, so t = 3.918964]

then plugging in t into y equation, we get y [use gravity in ft/s^2]

and then add 3ft b/c the ball was hit from 3 ft above the ground,

so the height = y + 3 = 148.35 + 3 = 151.35 ft