# (Physics) Finding velocity with angle and speed

Hi all, I'm very sorry if this is the wrong place to post a physics problem, but considering it's all maths I was hoping someone could help me. It's been a long time since I did maths on this or a similar level (nearly a decade) but I am learning physics for an exam I am planning to take this year and I have a long way to go!

I am trying to solve the a) calculation in the image.

I managed to work out the vertial (y axis) calculation. By doing: tan(30) on the calculator which gave 0.58 0.58*200m/s = 115m/s

Now I have tried to do the a) calculation and I know the answer should be 231m/s as it's in the back of the book, but I don't know how to get it. The closest I've got so far is 221ms. I have been using the Khan Academy video to help me but I can't get it right - I'm so tired from trying... http://www.youtube.com/watch?v=ZZ39o1rAZWY&list=SPAD5B880806EBE0A4

Can someone explain to me clearly why it is 231m/s. I tried to figure out the time and distance like he does in the video but I wasn't getting the right answer.

Thanks!

• Two ways: Let $s$ be the initial speed. Then $s\cos(30^{\circ})=200$. So $s=\frac{200}{\cos(30^\circ)}$. Or else find the answer to b) like you did. Then use Pythagorean Theorem. If the answer to b) is $B$, the answer to a) is $\sqrt{200^2+B^2}$. – André Nicolas Dec 20 '13 at 0:46
• Thank you! I think I overthought the equation and started adding in elements I didn't need... – Oli Dec 20 '13 at 22:47
• You are welcome. We have to keep the concrete situation constantly in mind, After that comes the manipulation. – André Nicolas Dec 20 '13 at 22:51

## 2 Answers

From Pythagoras $\sqrt{115^2+200^2}\approx 231$ Similarly, $\frac {200}{\cos 30^\circ}\approx 231$

• Thank you as well :), such a simple solution! – Oli Dec 20 '13 at 22:48

Hey I got the answer of actual velocity of missile. Just use Pythagoras theorem in the triangle. Let velocity of missile be $x$,so $$X^2=(200)^2+(115)^2$$ $$X^2=40000+13225$$ $$X^2=53225$$ $$X =230.7054$$ $$X = 231 m/s$$

• but this answer is already given. – A---B Aug 3 '16 at 14:57