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idku
 one year ago
I have another Q  harder than before:(
idku
 one year ago
I have another Q  harder than before:(

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idku
 one year ago
Best ResponseYou've already chosen the best response.0Ted 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
 one year ago
Best ResponseYou've already chosen the best response.0Can someone "walk" me step by step?

idku
 one year ago
Best ResponseYou've already chosen the best response.0They also gave me, just in case: 9.8 m/s^2 = 32.2 ft/s^2 1 mile = 5280 ft

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice drawing by the way ( the batter) lol

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440957695997:dw This is supposed to be a battler fighting a dragon

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440957532068:dw
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