A dive bomber has a velocity of 235 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.17 km. Find the angle θ.
To find the angle θ, we can use the equations of projectile motion. We can break down the motion of the bomb into horizontal and vertical components.
Let's start by considering the horizontal motion. The horizontal component of the velocity remains constant throughout the motion. We can find the time taken by the bomb to travel from the point of release to the target using the horizontal displacement.
Given:
Horizontal displacement (x) = 3.17 km = 3170 m
Horizontal velocity (Vx) = 235 m/s
We can use the equation:
x = Vx * t
Solving for t, we have:
t = x / Vx
t = 3170 / 235
t ≈ 13.51 s
Now that we know the time taken, let's consider the vertical motion of the bomb. We can find the initial vertical velocity (Vy) at the point of release using the formula:
Vy = V * sin(θ)
Given:
Initial altitude = 2.15 km = 2150 m
Vertical displacement (y) = 0 m (the bomb hits the ground)
Using the vertical motion equation:
y = Viy * t + (1/2) * g * t^2
Since y = 0 and the bomb is at its highest point at 2.15 km altitude, we can write:
y = Viy * t - (1/2) * g * t^2
Substituting for Viy and rearranging the equation:
2150 = V * sin(θ) * 13.51 - (1/2) * (9.8) * (13.51)^2
Now, we have one equation with one unknown (θ) that we can solve to find the angle:
2150 = 235 * sin(θ) * 13.51 - 4.9 * (13.51)^2
Simplifying the equation further, we have:
2150 = 3180.85 * sin(θ) - 884.89
Rearranging the equation:
3180.85 * sin(θ) = 3034.89
Dividing both sides by 3180.85:
sin(θ) = 0.9547
Now, to find the angle θ, we can take the inverse sine (sin^-1) of 0.9547:
θ = sin^-1(0.9547)
θ ≈ 72.44 degrees
Therefore, the angle θ is approximately 72.44 degrees below the horizontal.