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- idku

I have another Q - harder than before-:(

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- idku

I have another Q - harder than before-:(

- jamiebookeater

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- anonymous

what is it?

- idku

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!"
|dw:1440953349077:dw|
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?

- idku

Can someone "walk" me step by step?

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- idku

They also gave me, just in case:
9.8 m/s^2 = 32.2 ft/s^2
1 mile = 5280 ft

- anonymous

nice drawing by the way ( the batter) lol

- mathmate

Hint:
1. first split the velocity 176 ft/s into horizon and vertical components.
vx=176 cos(35)
vy=176 sin(35)
2. Ignoring air resistance,
find the time t it takes to reach the stadium wall.
x(t)=vx*t, where x=565, vx=176cos(34).
3. Find the height of the ball according to the kinematics equation:
y(t)=3+vy*t+(1/2)gt^2 (g=acceleration due to gravity = -32.2 ft/s^2, downwards).
the 3 ft. at the beginning because the ball is hit at 3 ft above ground.
4. y(t) is maximum height that the ball will clear. (I get over 150 ft.)

- Empty

|dw:1440957695997:dw| This is supposed to be a battler fighting a dragon

- mathmate

:) nice dragon!

- triciaal

|dw:1440957532068:dw|

- idku

thanks

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