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Physics 16 Online
OpenStudy (anonymous):

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 5.4 m/s at an angle of 13.6° below the horizontal. It is released 0.77 m above the floor. What horizontal distance does the ball cover before bouncing?

OpenStudy (anonymous):

\[H=(v^2\sin^2\Theta)/2g\]

OpenStudy (anonymous):

maybe that will help?

OpenStudy (anonymous):

english please. lol

OpenStudy (anonymous):

plug in the speed for v and the angle for sin gravity is 9.8 (g)

OpenStudy (anonymous):

may or may not help but it's worth a shot ha

OpenStudy (anonymous):

would it be 0.77=[(5.4)^2*(sin13.6)^2/2(9.8)? nice pun ;P

OpenStudy (anonymous):

lol wow ya idk if that's right O.o but hey I tried :D

OpenStudy (anonymous):

-.- RawR.

OpenStudy (anonymous):

sorry :)

OpenStudy (anonymous):

i guess nobody knows how to solve this...

OpenStudy (***[isuru]***):

Hey, 5.4 m/s = speed or velocity ?

OpenStudy (anonymous):

|dw:1379202463784:dw| initial speed

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