# Kinematics question

1. Sep 14, 2013

### GuN

1. The problem statement, all variables and given/known data
You throw a snowball with an initial speed of 15.0 m/s. Your enemy's head is located 20m away, approximately horizontally. At what angle should you throw the snowball (ignore air resistance)?

2. Relevant equations

vf=vi+at

d=vit+(1/2)at^2

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 15, 2013

### Tsunoyukami

Before we can help you we need to see your attempt at a solution. It might be useful for you to work through the problem in the following manner:

1) Make a list of all the different variables and their given values that you have.
2) Write out what it is the question is asking you to find.
3) Write out any equations you think will be useful for solving this problem.
4) Solve the problem using the equations above (this is often easier said than done!).
5) Write a final statement answering the question.

One hint I can give you is that motion in the x- and y- directions are independent of one another. Hopefully this helps you in your attempt at the problem!

3. Sep 15, 2013

### GuN

Here's what I got.

v initial (x component)=15.0 m/s v (y component)=0

d (x)=20m a (y)=-9.81 m/s^2

a (x)= 0 m/s^2

d (x)=v initial* t

t=d/v initial

=20/15

=1.33 seconds

v final (x component)= v initial (x) +a(x)t

but since a (x)=o, v final (x)= v initial (x)

so v final (x)=15.0 m/s

I think I remember my teacher saying x and y components both have the same time so:

v final (y)= v initial (y) + a (y) t

but if v intial (y) is zero v final (y)= a (y) t = (-9.81)(1.33)=-13.08 m/s or 13.08 m/s (down)

Then I made it into a right triangle, with the x component being 15.0 m/s and the y component being 13.08 m/s.

I did tan inverse of 13.08/15 being being 41 degrees.

I'm pretty sure this is wrong, because all of my friends and me are getting completely wrong answers, and the teacher's gone on sick leave (without providing answers to the practice assignment), the substitute has no clue what's she doing and the exam based on the homework is in a couple of days.

4. Sep 16, 2013

### Tsunoyukami

This question is not worded precisely and therefore we have to be careful in interpreting what it wants us to do. You have written
"at what angle should you throw the snowball" so we know we're interested in finding the angle of release of the snowball. Now let's think - what happens if you change this angle? Say $\theta$ = 0 corresponds to throwing the ball in the horizontal direction (ie. giving the snowball only a horizontal component). What happens if you throw the ball at an angle or 30 degrees? How about an angle of -30 degrees? HOw about an angle of 90 degrees? I recommend sketching each of the paths the snowball would take if thrown at these angles (and more if you're interested).

Next, based on the style of question and the inherent ambiguity, I will assume that you are probably in high school (maybe first year of university). If this is true the question likely means "at what angle should you throw the snowball" in order to hit your enemy in the head?

I'm glad you made an effort to follow my advised method (I really think it's useful especially when approaching new problems) but you've made a mistake - and that's okay. You should try tracing out the different paths a snowball would take when thrown at different angles.

Hint: Remember that the question provides the speed at which the ball is thrown but doesn't tell us in which direction (that's what we want to find out!) - you might find it useful to remember that you can decompose a vector into it's orthogonal components, that is, you can break down your velocity vector into x and y components. Do you know how to do this? (I'm pretty sure you do - hint: it involves using sin and cos)