A 60 mm mortar round is capable of being launched at 540 km/hr. If a target is located at a distance of 0.6 km and the mortar takes 7.5 s to reach the target, at what angle must the launcher be set to hit the target?
To find the angle at which the launcher must be set, we can use the equations of motion for projectile motion. We know that the time of flight, t, is 7.5 seconds, the initial velocity, vₒ, is 540 km/hr, and the horizontal distance, d, is 0.6 km.
First, we need to convert the initial velocity from km/hr to m/s. We know that 1 km/hr is equal to 0.2778 m/s. Therefore, the initial velocity is:
vₒ = 540 km/hr * 0.2778 m/s = 149.94 m/s
Now, let's break down the initial velocity into its horizontal and vertical components. We can use trigonometry to find these components. Let's assume the angle at which the launcher must be set is θ.
The horizontal component, vₒx, can be calculated by multiplying the initial velocity by the cosine of the launch angle:
vₒx = vₒ * cos(θ)
The vertical component, vₒy, can be calculated by multiplying the initial velocity by the sine of the launch angle:
vₒy = vₒ * sin(θ)
Now, let's calculate the time to reach the target using the horizontal component:
d = vₒx * t
Solving for vₒx:
vₒx = d / t = 0.6 km / 7.5 s = 0.08 km/s * 0.2778 m/s = 22.13 m/s
We can substitute this value back into the equation for the horizontal component:
vₒx = vₒ * cos(θ) = 149.94 m/s * cos(θ) = 22.13 m/s
Dividing both sides by vₒ:
cos(θ) = 22.13 m/s / 149.94 m/s ≈ 0.1476
Now, we can calculate the launch angle θ by taking the inverse cosine (or arccosine) of both sides:
θ ≈ cos⁻¹(0.1476) ≈ 81.85 degrees
Therefore, the launcher must be set at an angle of approximately 81.85 degrees to hit the target located at a distance of 0.6 km.