# find initial speed

• Sep 16th 2008, 01:40 PM
viet
find initial speed
Quote:

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

need help solving for $\displaystyle V_o$
• Sep 16th 2008, 01:44 PM
mr fantastic
Quote:

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

need help solving for $\displaystyle V_o$

Cancel $\displaystyle V_0$ in the first term on the right hand side. Note that $\displaystyle \frac{\sin 35^0}{\cos 35^0} = \tan 35^0$, which simplifies that term even more. Now make $\displaystyle V_0^2$ the subject ....
• Sep 16th 2008, 01:52 PM
viet
after cancelling $\displaystyle V_o$ i got
$\displaystyle 1.05 = tan(35)(13) - 4.9 (\frac{13}{V_o cos35})^2$

my algebra is a little rusty, can you show me how to make $\displaystyle V_o^2$ the subject?
• Sep 16th 2008, 02:06 PM
mr fantastic
Quote:

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

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

First isolate the second term by shifting all the constants to the left hand side.
• Sep 16th 2008, 02:30 PM
viet
let me know if im doing this wrong:
$\displaystyle 1.05 = 9.10-4.9(\frac{13}{V_o cos(35)})^2$

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

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

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

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

$\displaystyle V_o^2 cos(35) = 167.36$

$\displaystyle V_o^2 = 204.309$

$\displaystyle V_o = \sqrt{204.309} = 14.29$
• Sep 16th 2008, 05:36 PM
mr fantastic
Quote:

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

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

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

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

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

$\displaystyle V_o^2 cos(35) = 167.36$

$\displaystyle V_o^2 = 204.309$

$\displaystyle V_o = \sqrt{204.309} = 14.29$

..