Solve this physics problem!

He wouldn't aim above the sweet spot. The shot would take a little less than a second.

Physics Probs wrote:

Our HS physics teacher made up this problem and now nobody's sure if we have the right answer, including him.

A man is on a 50m high bluff and sees a buck in a clearing below. The buck stops 300m away from him (hypotenuse, not x distance). Assuming the sweet spot on the buck is 1.5 meters above the ground and his gun shoots at 300m/s NOT horizontally, how long does it take the bullet to hit the buck, and where would he have to aim in terms of meters above the sweet spot of the buck to hit the buck?

Since everyone on letsrun graduated from Harvard and now works as a doctor or astrophysicist I assumed this would be a great place to ask.

x = v0 cos theta t

y = v0 sin theta t - 1/2 gt^2

x = 295.8, y = -48.5, v0 = 300

Two solutions to this.

1) t = 0.996, theta = 0.146. This corresponds to aiming 92m above the buck's head.

2) t = 61.38, theta = 1.55. This corresponds to an almost-vertical shot aiming 18424m above the buck's head. I'd be very impressed if you could make this shot!

This problem does have a solution here is how you solve it:

1- Draw a diagram.

2- The variable you are solving for is the launch angle (theta).

3 - Split the problem into two parts, a normal projectile launch and a projectile thrown at a downward angle (same as the launch angle).

4 - You can get the vertical distance traveled = ht of cliff - ht of the buck

5 - You can get the total horizontal distance traveled from the Pythagorean theorem (diagonal is given).

6- The above = horizontal range of part I + horizontal range of II.

7- Solve for theta.

8- Once you have theta, draw a vertical line from the buck's head. Connect the launch velocity vector line to it to make a huge triangle.

9- Evaluate the height of this triangle.

10- Add it to the vertical distance in step 4. Done!

Easier (?) solution:

Just use the trajectory equation for a parabola:

y=xtan(theta)- gx^2/2v^2 cos^2(theta).

You have x and y, find theta (you have to do some trig gymnastics with half angles etc).

I was watching from the roof one afternoon when a group of roughly sixteen fully armed insurgents emerged from cover. They were wearing full body armor and were heavily geared. (We found out later that they were Tunisians, apparently recruited by one of the militant groups to fight against Americans in Iraq.) Not unusual at all, except for the fact that they were also carrying four very large and colorful beach balls. I couldn't really believe what I was seeing—they split up into groups and got into the water, four men per beach ball. Then, using the beach balls to keep them afloat, they began paddling across. It was my job not to let that happen, but that didn't necessarily mean I had to shoot each one of them. Hell, I had to conserve ammo for future engagements. I shot the first beach ball. The four men began flailing for the other three balls. Snap. I shot beach ball number two. It was kind of fun. Hell—it was a lot of fun. The insurgents were fighting among themselves, their ingenious plan to kill Americans now turned against them. “Y’all gotta see this,” I told the Marines as I shot beach ball number three. They came over to the side of the roof and watched as the insurgents fought among themselves for the last beach ball. The ones who couldn’t grab on promptly sank and drowned. I watched them fight for a while longer, then shot the last ball. The Marines put the rest of the insurgents out of their misery.

you have a sign error, as theta in solution 1 should be negative

also, i have:

x = 296.05, and y = -48.5

so, theta = -0.1463 (precision counts in 1000 foot shots), and t = .9975. So the shooter should aim 4.77m above the sweet spot.

note that 300m/s corresponds with a .22 rifle (perhaps an odd choice forsuch a long shot). A proper hunting rifle should be more like 800-900 m/s at the muzzle, which would mean the shooter would only need to aim something like 1.6m above the sweet spot.

also, note that we've been disregarding wind resistance, which gives us the odd result of t

What is the velocity of the round fired? What are the weather conditions?

Did you even check to see if your answer made sense before you posted?

This is what I got. I'll show my work in case I made a math error.

I tried to simplified as much as possible - we're just trying to kill a deer, not land on the moon.

x = horiz. distance from shot to target (m)

y = vert. distance from shot to target (m), we will make down positive

s = vertical distance traveled by the bullet (m)

t = time from shot to impact (s)

v = initial velocity = 300 m/s

u = initial vertical velocity (m/s)

g = acceleration due to gravity ~9.807 m/s^2

I have chosen to use an iterative solution, though there are other methods that could be used.

First we will determine how much we miss by if we aim directly at the target.

we will use the equation

s = u t + 0.5 a t^2

to determine how far our bullet has traveled down by the time it reaches the target.

gravity is the only force acting on the bullet (we can neglect drag), so a = g

***

to find t, we use

t = d / v

t = 300m / 300 m/s = 1s

to find u we use

u = v sin(angle of gun with horiz.)

we don't need to know the angle (though we can calculate it easily) since we have a right triangle with hypotenuse = 300, y = 48.5 (50-1.5), x = 295.804 (from Pythagorean theorem)

u = 300*(48.5/300) = 48.5 m/s

now we can solve

s = u t + 0.5 a t^2

s = 48.5*1 + 0.5*9.807*1^2 = 53.4035 m

so we shot 4.9035m too low (53.4035 - 48.5 = -4.9035)

now we start back from *** above, except not we aim 4.9035m higher.

this changes y to 43.5965m, x remains unchanged, the hypotenuse is now 298.999m

to find t, we use

t = d / v

t = 298.999m / 300 m/s = 0.997s

to find u we use

u = v sin(angle of gun with horiz.)

we don't need to know the angle (though we can calculate it easily) since we have a right triangle with hypotenuse = 298.999, y = 43.5965, x = 295.804

u = 300*(43.5965/298.999) = 43.742 m/s

now we can solve

s = u t + 0.5 a t^2

s = 43.742*0.997 + 0.5*9.807*0.997^2 = 48.4849 m

we wanted it to go 48.5 meters below us, so we missed by 1.5 cm

so that tells us we should aim somewhere between 4.9035m and 4.9185m above the what we want to hit. I think this is a small enough window, but if you want more accuracy, you can do another iteration.

Thank you for the responses! This one really cleared it up for me, thanks.