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.
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!