the horizontal velocity of the ball is
___ Vh = 176 cos(35º)
the vertical velocity of the ball is
___ Vv = 176 sin(35º)
it takes the ball 565 / Vh
seconds to reach the back wall
the time to the peak of the ball's
trajectory is ___ Vv / g
the difference between the peak time and the flight time to the wall is the time the ball has to fall from the peak height
___ this will give you the height at the wall
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?
2 answers
equations:
y= 0 + v*sin(35)*t - 1/2*g*t^2
x= v*cos(35)*t
[we have x=565, v=176 ft/s, so t=3.918964]
plug t into y equation, we get y [use gravity in ft/s^2]
and then add 3ft b/c the ball was hit from 3ft above the ground,
so the height = y+3 = 148.35 + 3 = 151.35 ft
y= 0 + v*sin(35)*t - 1/2*g*t^2
x= v*cos(35)*t
[we have x=565, v=176 ft/s, so t=3.918964]
plug t into y equation, we get y [use gravity in ft/s^2]
and then add 3ft b/c the ball was hit from 3ft above the ground,
so the height = y+3 = 148.35 + 3 = 151.35 ft
1 answer