Posted by Anonymous on .
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 = 0.56 m, and x = 7.0 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 time that the bullet spends in the building
t=x/v(0x) =7/340 =0.02 s.
The vertical displacement of the bullet in the building y= 0.56 m.
The vertical component of the velocity of the bullet as it passes through the window is
v(y) = (y- gt²/2)/t = y/t - gt/2=
=0.5/0.02 -9.8•0.02/2 =24.9 m/s
y=v(y)²/2g= 24.9²/2•9.8 = 31.6 m
H=31.6+0.56 =32.16 m
The time for the bullet to reach the window is
t₀ =2y/v(y) = 2•31.6/24.9 = 2.54 s
D=v(0x)t₀ = 340•2.54= 863.6 m
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