# Projectile motion, finding initial velocity

1. Sep 29, 2008

### fishingspree2

1. The problem statement, all variables and given/known data
A rock is thrown from a 1 meter height with a 35 degrees angle with horizontal. At x=64 m, its height is y=29 m. Find it's initial velocity.

2. Relevant equations

$$x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}$$
$$y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}$$

3. The attempt at a solution
Let F = initial velocity
then Vx0 = F cos 35
Vy0 = F sin 35
$$x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}$$
$$64=F\cos \left( 35 \right) t$$
$$t=64\,{\frac {1}{F\cos \left( 35 \right) }}$$

$$y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}$$
$$29=1+64\,\tan \left( 35 \right) - 10035.20000\,{\frac {1}{{F}^{2} \left( \cos \left( 35 \right) \right) ^{2}}}$$
solving for F gives 102 m/s
the answer in the book is 42 m/s

sorry i am new to latex

2. Sep 29, 2008

### alphysicist

Hi fishingspree2,

The number 10035.2 in the above equation does not look right to me.

3. Sep 29, 2008

### fishingspree2

sorry the calculation was made using 20070.4 i don't know how the 10035.2 showed up

i get 102m/s using 20070.4 and it is not correct according to the book

4. Sep 29, 2008

### alphysicist

I'm not getting 102m/s using 20070.4; can you show the details about how you get 102m/s?

5. Sep 29, 2008

### Integral

Staff Emeritus
I sorted out the constant. Thanks alphysicist.

It is usually a good idea to keep your expressions symbolic until the very end.

Show us how you are getting 102m/s.