Help with this problem : Projectile Motion

[MakeItThrough]

Homework Statement

http://www.webassign.net/CJ/3-45alt.gif ... (y = 0.46 m, and x = 6.6 m.)
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.46 m, and x = 6.6 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

Homework Equations

The basic equations for projectile motion:
Dx = v0x(t) + .5at^2
Dy = v0y(t) + .5gt^2

vx = v0x + at
vy = v0y +gt

vx^2 = v0x^2 + 2aDx
vy^2 = v0y^2 + 2gDy or as I like to call it 2gAy (for memorization purposes)
the D in these equations stand for distance

The Attempt at a Solution

My teacher did not explain this problem to us in class. I have no idea how to approach this problem.

*thanks for your response. I will try to solve this tomorrow and let you know if I have any problems then. Off to bed!

 

[Rayquesto]

2gAy? XD that's a really funny way to memorize. I love it!

ok so the best way to approach this problem is that you know the initial velocity of the launch in the y direction is zero. so, H=-1/2gt^2 so, t=sqrt2H/g

also vx is constant. so, Vxt=D + x

and it would probably be a good idea to find the velocity of the bullet when it goes throuugh the window.

so, here's my logic:
340m/s t=6.6meters when?

well the obvious answer is that it takes the bullet .019 seconds to go from the window to the wall.

next thing is that you know that y=.46meters and that it takes .019seconds to descend to the wall. so,

.46meters=(v0).019 + 1/2g(.019)

basicallly, you want to know the velocity in the y direction when it goes through the window and this is how you find it since V0 of this equation represents that.

once you find that, you can solve for how long it takes to get to that amount of y velocity by saying

velocity of the bullet in the y direction when it goes through the window=-gt and solving for t

once you know t you can solve for H and D easily. Let me know if you have problems. This question is complicated.

 

[MakeItThrough]

Thanks for your help! I got it!
There are two values for t. One of them is the motion of the bullet to the window. The other is the motion to the wall that the bullet hit.
what I did was set t = D/340 for motion to the window.
Then I made an equation regarding H and plugged in the value for t.
H - 0.46 = 0 + 4.9t^2
H = 4.9t^2 + 0.46
H = 4.9[(D + 6.6)/340]^2 + 0.46
H = (4.9D^2 + 53176) / 340^2 ... (1)

The equation for the other time: D + 6.6 = 340t ... t = (D + 6.6)/340
The 2nd equation for H:
H = 0 + 4.9t^2
H = 0 + 4.9[(D + 6.6)/340]^2
H = [4.9(D + 6.6)^2] / 340^2 ... (2)

Set equation 1 equal to 2 and I got D = 818.84 m

Next plug D into equation 1 and I got H = 28.88 m

 

[Rayquesto]

"Thanks for your help! I got it!
There are two values for t. One of them is the motion of the bullet to the window. The other is the motion to the wall that the bullet hit.
what I did was set t = D/340 for motion to the window.
Then I made an equation regarding H and plugged in the value for t.
H - 0.46 = 0 + 4.9t^2
H = 4.9t^2 + 0.46
H = 4.9[(D + 6.6)/340]^2 + 0.46
H = (4.9D^2 + 53176) / 340^2 ... (1)"

I think your algebra is a bit off right here.

(D + 6.6)^2=D^2 + 2(6.6)D + (6.6)^2

But you are definitely on the right path! :)

 

[rwisz]

*

I can offer some guidance on how to approach this problem. First, it is important to understand the concept of projectile motion and the equations involved. From the given information, we know that the bullet is moving horizontally at a constant velocity of 340 m/s and vertically due to the force of gravity. We also know that the bullet does not slow down as it passes through the window.

To solve for the distances D and H, we can use the equations for projectile motion. Since we are only interested in the horizontal distance (x), we can use the equation Dx = v0x(t) + .5at^2. We know that the initial horizontal velocity (v0x) is 340 m/s and the acceleration (a) is 0, since there is no horizontal force acting on the bullet. We can also assume that the time (t) it takes for the bullet to travel from the gun to the window is the same as the time it takes to travel from the window to the wall.

To find the time (t), we can use the equation t = Dx/v0x. Substituting the given values, we get t = 6.6 m/340 m/s = 0.0194 s. Now, we can use this value of t in the equation for vertical distance (y), Dy = v0y(t) + .5gt^2. Since we know that the bullet hits the wall at a height of 0.46 m, we can set up the equation as 0.46 m = v0y(0.0194 s) + .5(-9.8 m/s^2)(0.0194 s)^2. Solving for v0y, we get v0y = 142.8 m/s.

Now, we can use the equation vx^2 = v0x^2 + 2aDx to solve for the distance D. Again, since there is no horizontal acceleration, the equation becomes Dx = v0x^2/2a. Substituting the known values, we get Dx = (340 m/s)^2/(2*0) = 578400 m. This is the distance the bullet traveled horizontally from the gun to the wall. Therefore, the distance D is 578400 m - 6.6 m = 578393.4 m.

To find the height (H) of the building where