# Finding magnitude of velocity

1. May 5, 2013

### Suprin

1. The problem statement, all variables and given/known data
A ball is thrown upward from a platform 5.2meters high with a speed of 15m/s at an angle of 40° from the horizontal. What is the magnitude of its velocity when it hits the ground?

distance = 5.2m
$V_0$ = 15m/s
angle = 40°

2. Relevant equations
Don't have the answer to this question. Want to make sure I got the right answer.

3. The attempt at a solution

$V_0x = 15cos40°$ = 11.491

$V_0y = 15sin40°$ = 9.642

$|V| = \sqrt{11.491^2 + 9.642^2}$ = 15m/s ???

2. May 5, 2013

### voko

You got the magnitude of initial velocity. You need to find the final velocity.

3. May 5, 2013

### Suprin

Final velocity on X is equivalent to the initial velocity on X, so that remains the same.

$V_fy^2 = (9.642)^2 + 2(9.81)(5.2)$

$V_fy^2 = 92.968 + 102.024$

$\sqrt{V_fy^2} = \sqrt{194.992}$ = 13.964

$|V| = \sqrt{11.491^2 + 13.964^2}$ = 18.084

Out of curiosity, did I get the magnitude of the initial velocity right?

Last edited: May 5, 2013
4. May 5, 2013

### voko

You can check the initial velocity yourself, it was given :)

Regarding your solution, it seems correct, but. You should not use $V_{0y}$ for the vertical component of final velocity. $V_0$ is already used to mean initial velocity.

Finally, you don't have to use conservation of energy just for the vertical components. You could just as well have used it for total velocity.

5. May 5, 2013

### Suprin

I literally write down things the same way the professor gave us on our formula sheets.

And you completely lost me on the whole conservation of energy thing. If anything, that hasn't been mentioned during class.

6. May 5, 2013

### voko

The index 0 usually means "initial". You are looking for final velocity, so 1 or f would be more appropriate there, giving $V_1$ or $V_f$, respectively.

As for the conservation of energy, you used that to obtain the vertical component of the final velocity. If that sounds strange to you, explain what formula and used, and why.

7. May 5, 2013

### Suprin

Ah, ignore the sub-zero thing. That was a typo. Not used to writing math formulas on the PC.

We haven't worked with conservation of energy yet, nor air resistance or anything like that. The formulas I used (and why) are the ones I have available, for now, to obtain the information that is asked of me to find. It's one of those cases where I have about 4 formulas (for example), and I'm asked to find Time and only one of them has Time involved in it; so I know that's the correct one.

8. May 5, 2013

### voko

Uhm, OK. That formula does not require conservation of energy for its derivation, it just happens to follow most naturally from that. Anyway, your result is correct.