A sniper fires a bullet at 120 m/s at 30° above the horizontal from the roof top of a 35 m high parking garage. If the bullet strikes the level ground beside the parking garage:

a) How long was the bullet in the air?

b) How far from the base of the parking garage did the bullet land?

c) At what angle did the bullet land?

See Fri,5-2-14,2:17 AM post.

To determine the horizontal distance the bullet travels before hitting the ground, we need to break down the initial velocity of the bullet into its horizontal and vertical components.

The horizontal component of velocity (Vx) can be calculated using the given initial velocity (120 m/s) multiplied by the cosine of the launch angle (30°):

Vx = 120 m/s * cos(30°)

Vx = 120 m/s * 0.866

Vx ≈ 103.92 m/s

The vertical component of velocity (Vy) can be calculated using the given initial velocity (120 m/s) multiplied by the sine of the launch angle (30°):

Vy = 120 m/s * sin(30°)

Vy = 120 m/s * 0.5

Vy = 60 m/s

Next, we can calculate the time it takes for the bullet to reach the ground by using the vertical component of velocity (Vy) and the height of the parking garage (35 m). We can use the kinematic equation:

h = Vy * t + (1/2) * g * t^2

Where:
h = height
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)

35 m = 60 m/s * t + (1/2) * 9.8 m/s^2 * t^2

Rearranging the equation, we have a quadratic equation:

4.9t^2 + 60t - 35 = 0

Solving this equation will give us two possible values for t, but we discard the negative value since time cannot be negative. Therefore, we get:

t ≈ 1.165 seconds

Now that we have the time it takes for the bullet to reach the ground, we can calculate the horizontal distance (d) using the horizontal component of velocity (Vx) and the time (t):

d = Vx * t

d ≈ 103.92 m/s * 1.165 s

d ≈ 121.1 meters

Therefore, the bullet will travel approximately 121.1 meters horizontally before hitting the ground beside the parking garage.