# Homework Help: Kinematics: Angled Projectile Launch

1. May 7, 2014

### SlooM

1. The problem statement, all variables and given/known data
Determine the initial velocity of the projectile as it comes out of a cannon at a 20° angle.

2. Relevant equations
d = (Vi)(t) + 1/2(a)(t)^2

3. The attempt at a solution
I've split the question into both x and y components and solved for both using the equation above. Although I'm honestly unsure if I'm doing this correctly, would like some verification on my answer and point out any mistakes. Much appreciated! (Image below)

http://imgur.com/pPWO2jG

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Last edited: May 7, 2014
2. May 7, 2014

### Sunil Simha

Hi SlooM,

Are you sure v0y is 0 in the first equation that you've used to find t?

3. May 7, 2014

### SlooM

Ah, I think that may have been my mistake which is kind of why I look at it and it doesn't make sense. I just put 0 for v0y and assumed it canceled that half of the equation. What would I put though, I feel like I don't have enough variables and am super confused.

4. May 7, 2014

### Bashkir

In addition to what Sunil has stated, there was also an algebra mistake at the end preventing you from getting the correct answer.

$$\cos\theta = \frac{v_x}{v_i}$$

to

$$(cos\theta)v_x = v_i$$

$$v_i =\frac{v_x}{cos\theta}$$

An easier way to do this problem it to remember that your x-component of velocity already has a cosine term in it. So once you solve for time you can simply say

$$x(t) = V_otcos\theta$$

Last edited: May 7, 2014
5. May 7, 2014

### SlooM

Oh! I can't believe I didn't see that. So I redid the algebra at the end and got 44.5 m/s but are my calculations in the previous x and y calculations correct? I keep second guessing myself :/

I'm a rookie at physics but trying to improve. I appreciate all of your help and it means a lot.

6. May 7, 2014

### Sunil Simha

Hey we are all rookies

We get two equations right?

$-1.5= v_{0y}t + \frac{g}{2}t^2$
$23=v_{0x}t$

All we have to do is eliminate t. Moreover, v0x and v0y are related. So, you should be able to find v0x explicitly (and hence v0y).

Could you check you answer again? I seem to be getting something different.