Question 280279
You fire a rifle at an angle of 45 degrees. Thus the initial horizontal and vertical velocities of your bullet are the same. Suppose they each equal 240 feet per second. Again ignore air resistance. Assume you are shooting from ground level (height 0). Your bullet will hit the ground ____  feet from your current position. 
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This is really a physics problem.
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Applying the equation of motion:
y = (1/2)(-g)t^2 + vt + yi
where
g is gravity 32 feet per second squareed
t is time in secs
v is initial velocity
yi is the initial height (0)
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0 = (1/2)(-32)t^2 + 240t + 0
0 = -16t^2 + 240t
0 = -t^2 + 15t
0 = t^2 - 15t
0 = t(t - 15)
t = 15 seconds (time it will hit ground)
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equation of motion in the x direction:
x = (1/2)at^2 + vt + xi
Since acceleration in the x direction is zero
and, initial start is zero we have
x = vt
x = 240(15)
x = 3600 feet