Posted by **Jill** on Sunday, October 18, 2009 at 2:59pm.

A player is shooting the basketball 10 meters from the hoop. He releases the ball when it is 2 meters above the floor, with an initial velocity of 12 m/s at an angle of 65 degrees. ( the basketball hoop is also 3m above the ground)

How long does it take for the ball to reach the hoop?

What is the maximum heigh above the ground that the ball reaches?

How fast is the ball going when it is as its max height?

I have been having sooooo much trouble with this question.

- Physics -
**bobpursley**, Sunday, October 18, 2009 at 3:03pm
Start with the vertical equation.

hf=hi+12Sin65*time -4.9t^2 solve for time in the air.

Max height? When is vvertical zero?, then use that time to calculate vertical height in the first equation.

how fast at top? 12 Cos 65

- Physics -
**Phil**, Sunday, October 18, 2009 at 3:07pm
Ok so I got the first and last parts correct from what i did, but the second part is still messing me up. Do i enter 1/2 the time i calculated in the first question?

- Physics -
**bobpursley**, Sunday, October 18, 2009 at 3:15pm
No. refigure time to the top, vv=0

vfv=0=12Sin65 - 9.8t^2

solve for time t to the top.

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