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
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
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with the equations of motion:
x(t)=v0 cos(theta) t
y(t)=y0 + v0 cos(theta) t - 1/2 gt^2
use the x equation to get t=x/v0 cos(theta)
substitute this into the y equation and get
y(x)= y0 + x tan(theta) - gx^2/(2 v0^2 cos^2(theta))
we want to find the value of y when x=565 ft, v0=176ft/s, theta = 35, and g=32.2 ft/s/s, and y0=3 ft
we have
y(x) = 3+ 565 tan 35 - 32.2 (565)^2/(2*176*cos35)^2
y(x) = 274 ft
this is the height of the needed fence (without friction, a baseball hit with these parameters would travel over 900 feet)
x(t)=v0 cos(theta) t
y(t)=y0 + v0 cos(theta) t - 1/2 gt^2
use the x equation to get t=x/v0 cos(theta)
substitute this into the y equation and get
y(x)= y0 + x tan(theta) - gx^2/(2 v0^2 cos^2(theta))
we want to find the value of y when x=565 ft, v0=176ft/s, theta = 35, and g=32.2 ft/s/s, and y0=3 ft
we have
y(x) = 3+ 565 tan 35 - 32.2 (565)^2/(2*176*cos35)^2
y(x) = 274 ft
this is the height of the needed fence (without friction, a baseball hit with these parameters would travel over 900 feet)