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?

Question

anonymous · Answer

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

anonymous · Answer

maybe that will help?

anonymous · Answer

english please.
lol

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

-.- RawR.

anonymous · Answer

sorry :)

anonymous · Answer

i guess nobody knows how to solve this...

***[isuru]*** · Answer

Hey,
   5.4 m/s = speed or velocity ?

anonymous · Answer

|dw:1379202463784:dw|
initial speed