Find Angle to Throw Snowball for 20m Distance | Kinematics

[GuN]

Homework Statement

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)?

Homework Equations

vf=vi+at

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

The Attempt at a Solution

 

[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!

 

[GuN]

Thanks for the advice.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 secondsv 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/sI 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.

 

[Tsunoyukami]

Homework Statement

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)?

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?

Thanks for the advice.
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 secondsv 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/sI 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.

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)

 

[Nickie]

To find the angle at which you should throw the snowball, we can use the kinematic equations. First, we need to find the time it takes for the snowball to travel 20m horizontally. Using the equation d=vit+(1/2)at^2, we can rearrange it to solve for time (t):

t = (2d/vi)^0.5

Plugging in the given values, we get t = (2*20m/15.0 m/s)^0.5 = 1.825 seconds.

Now, using the equation vf=vi+at, we can find the vertical component of the snowball's velocity at the time it reaches the 20m mark. Since we are ignoring air resistance, the vertical component of velocity will be constant throughout the motion. So we can set vf=0, vi=0, and a=-9.8 m/s^2 (acceleration due to gravity). Solving for vi, we get vi=9.8 m/s.

Next, we can use the trigonometric relation tan(theta) = opposite/adjacent to find the angle at which the snowball should be thrown. In this case, the opposite side is the vertical component of velocity (9.8 m/s) and the adjacent side is the initial velocity (15.0 m/s). So, tan(theta) = 9.8/15.0 = 0.653. Taking the inverse tangent of both sides, we get theta = 33.6 degrees.

Therefore, to throw a snowball 20m horizontally with an initial speed of 15.0 m/s, you should throw it at an angle of 33.6 degrees above the horizontal.