Question

A bullet is fired at an angle of 60° with an initial velocity of 200.0 m/s. How long is the bullet in the air? What is the maximum height reached by the bullet?

Answer 1

To find the total time the bullet is in the air, we split the initial velocity into horizontal and vertical components:

Horizontal component: Vx = V * cos(theta) = 200.0 m/s * cos(60°) = 200.0 m/s * 0.5 = 100.0 m/s

Vertical component: Vy = V * sin(theta) = 200.0 m/s * sin(60°) = 200.0 m/s * 0.866 = 173.2 m/s

Next, we can calculate the time the bullet spends in the air by considering the vertical motion:

Using the equation h(t) = Vy*t - 0.5*g*t^2, where h(t) is the height at time t, Vy is the initial vertical velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time.

When the bullet reaches its maximum height, its vertical velocity becomes 0, so we set Vy - g*t = 0 and solve for t:

173.2 m/s - 9.81 m/s^2 * t = 0
9.81 m/s^2 * t = 173.2 m/s
t = 173.2 m/s / 9.81 m/s^2 ≈ 17.67 seconds

Since the bullet will be in the air for twice as long as it takes to reach maximum height before coming back down, the total time the bullet is in the air is approximately 35.34 seconds.

To find the maximum height reached by the bullet, we substitute the time t back into the equation for height:

h(t) = 173.2 m/s * 17.67 s - 0.5 * 9.81 m/s^2 * (17.67 s)^2
h(t) = 3063.2 m - 0.5 * 9.81 m/s^2 * 313.56 s^2
h(t) = 3063.2 m - 1538.68 m
h(t) ≈ 1524.52 meters

Therefore, the bullet is in the air for approximately 35.34 seconds and reaches a maximum height of approximately 1524.52 meters.