Below A gun is fired parallel to the ground. As the bullet leaves the barrel of

Question

Below

A gun is fired parallel to the ground. As the bullet leaves the barrel of the gun with a velocity of 200 m/s, find the following (neglecting air resistance): Velocity in the x direction (vx) Acceleration in the x direction (ax) Velocity in the y direction (vy) Acceleration in the y direction (ay) Assuming the barrel is 50m above the ground, calculate how much time passes before the bullet hits the ground. Using your answers above, calculate how far the bullet travels in the x- direction. The same gun is fired at a 65 degree angle with respect to the horizontal. Answer parts a-f from question one above.

Explanation / Answer

1(a) 200m/s
(b) 0
(c) 0
(d) 9.8 m/s^2
(e) By s = ut + 1/2gt^2
=>50 = 0 + 1/2 x 9.8 x t^2
=>t = ?10.20
=>t = 3.19 sec
(f) By R = [Ux] x t
=>R = 200 x 3.19 = 638.88 m

2.65 degrees above or below the horizontal? I would assume above, it's a more difficult problem that way.

You just have to do the same thing as above, but calculate the velocity for x and y from initial v:
vx = cos(65) * 200 m/s = 84.5 m/s
vy = sin(65) * 200 m/s = 181 m/s

Acceleration is still 0 and -9.8 m/s^2.

2e has to be done in 2 parts - the way up to the peak, and back down. Since you know constant acceleration, you can calculate time for the bullet to stop:
t = vy / ay = 18.5 seconds
and x traveled here is 18.5 seconds * vx, or 1561 m.
On the way down, we know that when the bullet falls back to the height it started, it's going the same speed as it was initially, but in the opposite direction. It takes the same amount of time and travels the same x distance. So at that point, it still has another 50 m to go.

d = 1/2 * g * t^2 + v0 * t + x0
0 = -4.9 t^2 + -181 * t + 50
t = 0.27 s

Ttotal = 2*18.5 + 0.274 = 37.3 s

In that time, the bullet traveled 37.3 * 84.5 = 3152 m

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