## Snowball fight.

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the mishanator
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Joined: Mon May 31, 2010 6:49 pm UTC

### Snowball fight.

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, you throw a second one at a low angle and timed to arrive at your opponent before or at the same time as the first one. Assume both snowballs are thrown with a speed of 30.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal.

(a) At what angle should the second (low angle) snowball be thrown if it is to land at the same point as the first?

(b) How many seconds later should the second snowball be thrown if it is to land at the same time as the first?
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There are a number of things that I do not understand with this question. 1) How can they land at the same time if their velocities are the same but angles different? 2) In part (a) it does not explicitly state how much later the second snowball is thrown, so how can I come up with an answer? Doesn't any angle qualify as long as you throw it at the right time? 3) In part (b), the same type of concern: how can you have a time without an explicit angle!?

masher
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Joined: Tue Oct 23, 2007 11:07 pm UTC
Location: Melbourne, Australia

### Re: Snowball fight.

Quoted to unbeak the post...

the mishanator wrote:One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, you throw a second one at a low angle and timed to arrive at your opponent before or at the same time as the first one. Assume both snowballs are thrown with a speed of 30.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal.

(a) At what angle should the second (low angle) snowball be thrown if it is to land at the same point as the first?

(b) How many seconds later should the second snowball be thrown if it is to land at the same time as the first?
____________________________________________________________
There are a number of things that I do not understand with this question. 1) How can they land at the same time if their velocities are the same but angles different? 2) In part (a) it does not explicitly state how much later the second snowball is thrown, so how can I come up with an answer? Doesn't any angle qualify as long as you throw it at the right time? 3) In part (b), the same type of concern: how can you have a time without an explicit angle!?

1) They can land at the same time because they are thrown at different times.
2) a) is not asking about time, just position. Any angle doesn't qualify, as you're only allowed to thrown that snowball at 30m/s.
3) You've already calculated the distance you need to throw the second snowball (ie the same as the first). You know the velocity you're throwing it at (ie 30m/s). From these you can calculate an angle to be thrown at, and a time that it will take to get there.

a) Ignore the time aspect. Given that the first snowball is thrown at 30m/s at 75º, how far does it travel in the horizontal? Now that you know how far snowballs 1 & 2 travel, you should now be able to calculate the angle at which to throw the second snowball, given that you are also throwing it at 30 m/s.

b) Now that you know the angles and whatnot, how long does it take the first snowball to travel from your hand to your opponent's face? How long does it take the second? What is the difference between these two values? This is the time between throwing the first and second snowballs to ensure that they both land on his face at the same time.
Last edited by masher on Mon Oct 04, 2010 4:23 am UTC, edited 1 time in total.

the mishanator
Posts: 209
Joined: Mon May 31, 2010 6:49 pm UTC

### Re: Snowball fight.

well okay.
for (a) i have the equation [imath]y=y_0+v_{0y}t-\frac{1}{2}gt^2[/imath], filling it in i get (since [imath]y_0[/imath] is zero) [imath]y=30\cos{75}t - \frac{1}{2}\left(9.8\right)t^2[/imath] so i solve for [imath]t[/imath] and get that [imath]t \approx ..4517\left(\sqrt{42.8425-y}+6.5454\right)[/imath] but what exactly does this tell me? or am i not on the right track? and if it's right, how do i use it?

masher
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Joined: Tue Oct 23, 2007 11:07 pm UTC
Location: Melbourne, Australia

### Re: Snowball fight.

You're on the right track...

You don't need to know vertical displacements, only horizontal. So, to calculate the distance the first snowball travels:

Solve for t:
$v = u + at$
$0 = 30 \sin 75 + (-9.8) t$

That's the time to the apogee, so the total transit time from you hand to his face is [imath]2t[/imath]

Now that you know the total travel time, you can calculate the total horizontal distance the snowball travelled.

the mishanator
Posts: 209
Joined: Mon May 31, 2010 6:49 pm UTC

### Re: Snowball fight.

oh okay.
take 2:

given [imath]x=x_0+v_{0x}t[/imath], [imath]x = 30\cos{75}(2.95692) \approx 22.9592[/imath]

now what? i have the distance, but since it doesn't give the angle at which the ball is thrown, i can't really set up any sort of equality...
Last edited by the mishanator on Mon Oct 04, 2010 4:38 am UTC, edited 1 time in total.

masher
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Location: Melbourne, Australia

### Re: Snowball fight.

Third ball?

You're only throwing two.

the mishanator
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### Re: Snowball fight.

oops... i have no idea what i was thinking... that should just be "the ball" ill fix right now. sorry.

masher
Posts: 821
Joined: Tue Oct 23, 2007 11:07 pm UTC
Location: Melbourne, Australia

### Re: Snowball fight.

You're out by a factor of two (that t is the time to apogee, you need twice that to get back down to the ground). The distance the ball travels is 45.89 m.

This is the distance the first snowball travels. It is also the distance the second snowball travels.

Now, at what angle (other than 75º) can you throw a snowball at, with an initial velocity of 30 m/s, such that it travels 45.89 m horizontally?
Last edited by masher on Mon Oct 04, 2010 4:47 am UTC, edited 1 time in total.

the mishanator
Posts: 209
Joined: Mon May 31, 2010 6:49 pm UTC

### Re: Snowball fight.

im still not quite getting this.

i would have the equation: [imath]45.89 = \cos{\theta}t[/imath] am i assuming t is the same here? or what?

masher
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Joined: Tue Oct 23, 2007 11:07 pm UTC
Location: Melbourne, Australia

### Re: Snowball fight.

You've got:
$v=u+at$
which can be rearranged as (assuming [imath]v=0[/imath])
$t=\frac{u}{a}$
and then you can substitute [imath]u[/imath]:
$t=\frac{30\sin\theta}{9.8}$

The horizontal displacement is given by
$s_h = 2vt$
as there isn't any acceleration in the horizontal. The factor of 2 comes from the time calculation only going to apogee.

So it follows that the horizontal displacement can be expressed as
$s_h = 2 v \frac{30\sin\theta}{9.8}$
and then too
$45.89 = 2 \times 30\cos\theta \frac{30\sin\theta}{9.8}$

You've gotten rid of your time dependency and can now solve for [imath]\theta[/imath].

tdug1991
Posts: 8
Joined: Fri Oct 01, 2010 8:52 pm UTC

### Re: Snowball fight.

R = ( V^2*sin(2[theta]) )/g

with [theta] being the launch angle
R = range (distance traveled by projectile)
g = gravity (9.
V = launch velocity

the mishanator
Posts: 209
Joined: Mon May 31, 2010 6:49 pm UTC

### Re: Snowball fight.

heh, i figured it out. i used the 'y' equation to solve t, then the 'x' to solve for... well x. then set 'em equal and BAM!!!! bolts of lightening everywhere and i got the right answer. thanks for the help gentlemen (or women).

gorcee
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### Re: Snowball fight.

To give some context for this example in real life, modern computerized artillery batteries use this same concept to time the landing of (sometimes different types of) artillery shells. This is called Multiple Round, Simultaneous Impact (MRSI): http://en.wikipedia.org/wiki/Artillery#MRSI

the mishanator
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### Re: Snowball fight.

that happens to be the example the professor used in the lecture

masher
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Location: Melbourne, Australia

### Re: Snowball fight.

the mishanator wrote:thanks for the help gentlemen

No worries.