You decide to shoot your pellet gun at a target on the wall of your backyard. You are sitting in a tree, 5.60 m high. You are aiming directly at the center of the bull's eye, which is located 1.20 m above the ground. The tree is located 9.10 m away from the wall. Your pellet gun has a muzzle velocity of 13.5 m/s.

a) how long is the pellet in the air?
I got 1.55 s

b) does your pellet even reach the wall? If so, then determine how far up it hits and the velocity your pellet had when it struck. If not, determine how far from the wall it struck the ground, and with what landing velocity


c) at what angle with the horizon should you aim your pellet gun to hit the bull

These last two I don't know what to do

a) how long is the pellet in the air?
...
b) does your pellet even reach the wall? if so, then determine how far up it hits and the velocity your pellet had when it struck. If not, determine how far from the wall it struck the ground, and with what landing velocity
...
c) at what angle with the horizon should you aim your pellet gun to hit the bull


a) the time an object spends in the air is directly related to only its initial velocity in the y direction if you neglect air resistance and drag forces.

The formula I would use is

y_f=\frac{gt^2}{2}+V_{iy}t+y_0 Where yf is final horizontal position, Viy is initial horizontal velocity, g is the accereleration due to gravity, and y0 is initial horizontal position. The only unknown here is t.

Initial velocity in the horizontal direction can be found using the information given. We have a triangle with base 9.1m and height of 5.6-1.2(we subtract because the target is 1.2m off the ground! Note that this also means you are aiming downwards if you are pointed to the center of the target) thus the angle to the ground can be given if we use some of our trig rules.

\theta=arctan(\frac{-4.4}{9.1}) (The height is negative because you are at a position that is taller than the target)
\theta=-25.8 deg

Then to find initial horizontal velocity, we utilize again more trig

V_{0y}=13.5sin(-25.8) (13.5 is the initial velocity given to us, we want to split it up to components)
V_{0y}=-5.88 m/s

Now back to our original equation for horizontal velocity we have:
0=-4.9t^2-5.88t+5.6
I could be wrong, but you may want to double check your answer, I am getting a flight time of about 0.626 seconds

b)Now that we have determined flight time by plugging time into a similar function we used before but for horizontal position.

x_f=\frac{at^2}{2}+V_{ix}t+x_0 (acceleration in the x direction is 0)

Since we know the angle you are shooting at, we can find Vix easily using more trig.

V_{ix}=13.5cos(-25.8)
V_{ix}=12.12 m/s

Then we have

x_f=12.12*0.626 initial horizontal position, and acceleration is 0
x_f=7.61 m

It only travels 7.61 meters so its 1.49m short!

The landing velocity is the addition of the vectors in the x and y direction. We already know x. its 12.12m/s, the y direction can be found in a similar manner using

V_y=gt+V_{iy}
V_y=-4.9(0.626)-5.88
V_y=-8.95

Then us a little bit of pythagorean magic and we get

V_f=15.06m/s

c) I'll leave this one for you, Utilize all the tools above to answer this question! If you have any questions please ask. If I got the time wrong, then please show me my mistakes since this is a rather long problem. (though the question in part c makes it highly likely that the bullet does indeed not make its way to the target successfully with the given conditions)

Keep in mind a change in angle would mean a change in only the y component of your initial and final velocities!

a) Well, you did make a mistake, Nhatkiem. You took displacement as 5.6, though you said that you have to take 5.6-1.2, lol!

Using vertical component, I get 0.52 s...

You cannot use the horizontal component, since you don't know where the pellet will exactly land. If you use the horizontal component here, and assuming that there were no solid ground, the pellet would dig through the ground or wall by the time it reaches the 9.1 m. You'll see this in the next part.

b) Now that you have time, use the formula that Nahtkiem gave you, this time using horizontal component. So, you see that it does not reaches it's goal. But now, the pellet is still off the ground, 1.2 m high in the air at the time you used above! You now use the horizontal component to find the time.

You should get 0.75 s. [I saw that you used the wrong angle in your first part. Use a sketch with motion problems, they help a lot!]

Now, find the height of the bullet at that time, using the formula Nhatkiem gave you. You should get 7.15 m which is way underground!

So, another option, you have to find the distance from the wall. You have to find yet another time of flight, using the same formula, this time, the magnitude of displacement will be 5.6, the height from which you shoot. Now, you should have the time as 0.63 s. The distance the pellet travels is then 7.61 m. So your pellet travelled 7.61 m, which is 1.49 m away from the wall.

{Nhatkiem, it is there that your answer miraculously coincides lol! :p}

c) Ok, that one is lots tougher, but not impossible.You have to cover 9.1 m horizontally, but at the same time, 4.4 m downwards. The time for both cases must be the same.

s = u_ht for horizontal component. So, t = \frac{s}{u_h}

s = u_vt + \frac12 at^2 [my way of using the formula]. So, t =\frac{-u_v + \sqrt{u_v\ ^2 - 4(0.5a)(-s)}}{2(0.5a)}

Since the time is the same, equate both to get the initial vertical component.

\frac{s}{u_h} = \frac{-u_v + \sqrt{u_v\ ^2 - 4(0.5a)(-s)}}{2(0.5a)}

\frac{9.1}{u_h} = \frac{-u_v + \sqrt{u_v\ ^2 +86.38}}{9.81}

Note that I'm taking downwards as positive to remove all the negative values.

Now, u_h = 13.5cos theta
u_v = 13.5sin theta
where theta is the angle with the horizontal, and clockwise is positive since I took downwards as positive.

So;

\frac{9.1}{13.5cos\theta} = \frac{-13.5sin\theta + \sqrt{(13.5sin\theta)^2 +86.38}}{9.81}

\frac{9.1}{13.5cos\theta} = \frac{-13.5sin\theta + \sqrt{182.25sin^2\theta +86.38}}{9.81}

Well, that's awfully long... I'll give an answer later. I'll be trying to solve that graphically, using a software, phew!

a) Well, you did make a mistake, Nhatkiem. You took displacement as 5.6, though you said that you have to take 5.6-1.2, lol!
...
So, another option, you have to find the distance from the wall. You have to find yet another time of flight, using the same formula, this time, the magnitude of displacement will be 5.6, the height from which you shoot. Now, you should have the time as 0.63 s. The distance the pellet travels is then 7.61 m. So your pellet travelled 7.61 m, which is 1.49 m away from the wall.

{Nhatkiem, it is there that your answer miraculously coincides lol! :p}


Actually my work there is correct. At least from my point of view. I found the time it took for the bullet to travel 5.6 m downwards. The angle at which the gun was pointed is critical. The gun is pointed downwards, so there is an initial y component downwards. The triangle's height is 4.4, but the TOTAL horizontal displacement is 5.6. I used 4.4 to calculate the angle of initial trajectory, but I used 5.6 because that is overall displacement.

And you don't need to know if the bullet will actually land, you just need to know what the maximum time it will be in the air for since that is what ultimately determines whether the bullet lands.

It's not coincidence that we both ended up with ~0.626 seconds :cool:

Oh yes, it's only after trying c) in the software that I realised it... By the way, the answer I get for c) is weird... and I can't seem to find my mistake.

Oh yes, it's only after trying c) in the software that I realised it... By the way, the answer I get for c) is weird... and I can't seem to find my mistake.

I'll see if I can take a look at it after school. I just finally received my financial aid award letter after waiting for 9 months haha... Anyway I got some paper work to fill out and some of my homework to do, but I'm pretty sure whatever "weird" answer you came up with is correct :p

You sure? I plotted the graphs, and I have times of either 0.34 s with 0.05 rad or ~13 s with a greater angle. I don't think either is good... either there's some other way to find it or I missed something.

You sure? I plotted the graphs, and I have times of either 0.34 s with 0.05 rad or ~13 s with a greater angle. I don't think either is good... either there's some other way to find it or I missed something.

If you used quadratic to figure out t, did you also test for when it would be

b^2-expression and not just b^2+expression?

Anyway, I plugged in your equation hoping maple would be able to figure out theta algebraically and it froze on me haha, Let me try it graphically.

Lol Lol Lol! I didn't see that coming :p

Aaaahhh!

No, I didn't try because that would give a negative time and I don't think that we can have it the negative way around.

This is just one of those problems that's easy to set up.. but a pain in the butt to do the actual computation. Algebraically I do not see anything wrong with the way you have set up the problem.

But the number crunching isn't yielding any results. Trying to get a ratio of sines and cosines to equal each other usually ends up as a jumbled mess.

Yes, I saw it as soon as I got cos^2Asin^2A,cosAsinA and sin^3A. I first thought that a quadratic in sin would be okay, but then, cosAsinA showed up and blocked everything. :o