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
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.
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