find initial speed

**viet** · September 16th 2008, 01:40 PM

A basketball player who is 2.00 m tall is standing on the floor L = 13.0 m from the basket, as in the figure below. If he shoots the ball at a 35.0° angle with the horizontal, at what initial speed must he shoot the basketball so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.

so far i got:
$1.05 = V_o sin(35)(\frac{13}{V_o cos(35)}) + \frac{1}{2}(-9.8)(\frac{13}{V_o cos(35)})^2$

need help solving for $V_o$

**mr fantastic** · September 16th 2008, 01:44 PM

Originally Posted by viet

so far i got:
$1.05 = V_o sin(35)(\frac{13}{V_o cos(35)}) + \frac{1}{2}(-9.8)(\frac{13}{V_o cos(35)})^2$

need help solving for $V_o$

Cancel $V_0$ in the first term on the right hand side. Note that $\frac{\sin 35^0}{\cos 35^0} = \tan 35^0$ , which simplifies that term even more. Now make $V_0^2$ the subject ....

**viet** · September 16th 2008, 01:52 PM

after cancelling $V_o$ i got
$1.05 = tan(35)(13) - 4.9 (\frac{13}{V_o cos35})^2$

my algebra is a little rusty, can you show me how to make $V_o^2$ the subject?

**mr fantastic** · September 16th 2008, 02:06 PM

Originally Posted by viet

after cancelling $V_o$ i got
$1.05 = tan(35)(13) - 4.9 (\frac{13}{V_o cos35})^2$

my algebra is a little rusty, can you show me how to make $V_o^2$ the subject?

First isolate the second term by shifting all the constants to the left hand side.

**viet** · September 16th 2008, 02:30 PM

let me know if im doing this wrong:
$1.05 = 9.10-4.9(\frac{13}{V_o cos(35)})^2$

$-8.05 = -4.9(\frac{13}{V_o cos(35)})^2$

$1.64 = \frac{13^2}{(V_o cos(35))^2}$

$1.64 = \frac{168}{(V_o cos(35))^2}$

$V_o^2 cos(35)^2+1.64 = 169$

$V_o^2 cos(35) = 167.36$

$V_o^2 = 204.309$

$V_o = \sqrt{204.309} = 14.29$

**mr fantastic** · September 16th 2008, 05:36 PM

Originally Posted by viet

let me know if im doing this wrong:
$1.05 = 9.10-4.9(\frac{13}{V_o cos(35)})^2$

$-8.05 = -4.9(\frac{13}{V_o cos(35)})^2$

$1.64 = \frac{13^2}{(V_o cos(35))^2}$

$1.64 = \frac{168}{(V_o cos(35))^2}$ Mr F says: The 168 should be 169. Typo ....?

$V_o^2 cos(35)^2+1.64 = 169$ Mr F says: Should be ${\color{red}V_o^2 cos(35)^2 \times 1.64 = 169}$ . There will be a ripple effect - you'll need to make the necessary corrections below.

$V_o^2 cos(35) = 167.36$

$V_o^2 = 204.309$

$V_o = \sqrt{204.309} = 14.29$

..

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