# Kinematics in Two Dimensions

1. Mar 22, 2015

### kaspis245

1. The problem statement, all variables and given/known data
I need to find velocity v from the drawing.

2. Relevant equations
Equations of kinematics

3. The attempt at a solution

I found that v=12.134 m/s . I need to know is it correct.

2. Mar 22, 2015

### Delta²

Your mistake should be in the calculation of s2. Since you use the equation $h=1/2gt_{2}^2$ seems like you ignore the fact that the particle has vertical velocity $v_{y}=gt_{1}/2$ at that point. So it should be $h=v_{y}t_{2}+1/2gt_{2}^2$

3. Mar 22, 2015

### paisiello2

Maybe you can verbally explain your reasoning?

4. Mar 22, 2015

### BrettJimison

Delta means this:in order to find your time for s2, you find the time it takes to fall a distance h under gravity with an initial velocity (in the y direction) of zero. This will give you the wrong time. You need to account for the initial vertical velocity. Think: which takes longer to hit the ground: a ball shot straight forward or a ball shot upward at an initial positive angle above the horizontal?

5. Mar 22, 2015

### paisiello2

I think it's a mistake to combine s1 and s2 the way you are doing because they depend on t^2.

The standard way to solve a problem like this is to break into two parts: Vx and Vy and use t as the common variable.

6. Mar 23, 2015

### kaspis245

I understand my mistake, it should be h=vyt2+1/2gt22

Here's what I get:

Then I tried to solve it:

By the way, I did solve this equation using Wolfram Alfa and got v=6,88 m/s.

Last edited: Mar 23, 2015
7. Mar 23, 2015

### Delta²

You got drown in a spoon of water :D, you forgot the square root in the last step, the velocity is the square root of 47.4.

But i am not quite sure the initial equation ($sg=2v_x^2+...$ is correct can u show us how u derive it.

Ok i think it must be correct cause solving the problem with the "usual" aaproach described briefly by paisiello2 gives the same result.

Last edited: Mar 23, 2015
8. Mar 23, 2015

### ehild

There is a much simpler method based on the trajectory of the projectile. The horizontal displacement (x) is x = Vo cos(α) t , the vertical displacement y - h= Vosin(α) t -g/2 t2. Eliminate the time t, you get the equation y(x) for the whole trajectory:
$y = h + \tan(α) x - \frac{g x^2}{2 V_0^2 \cos^2(α)}$
You know both the initial point and the final point of the trajectory: if x=0, y=h=0.17 m and x=s=5 m, when y=0. Alpha is given, α=45°. Substitute the known data, and solve the equation for Vo.

Last edited: Mar 23, 2015