Question

A missile is launched vertically. A camera 1000m from the launch site is tracking the missile. If the camera is rotating upward at 0.05 radians/sec when the angle of elevation is 60 degrees, find the velocity of the missile.

Answer 1

The formula relating rocket height to camera angle is:

tan(θ) = h/1000, where h is the height of the rocket and θ is the camera angle.

This is a related rates problem.

Let's differentiate both sides of the equation by t (time).

sec^2(θ)*(dθ/dt) = (1/1000)(dh/dt)

We now have an equation relating the current angle and height, and the rates of change of angle and height.

dθ/dt represents the rate of change in angle per second, dh/dt represents the rate of change in height per second

Let's fill in all variables that we know.

θ = 60 deg = π/3 rad

dθ/dt = 0.05 rad/s

"Find the velocity of the missile", means finding the rate of change in height of the missile, so finding dh/dt.

Let's plug in the values for θ and dθ/dt, and solve for dh/dt.

sec^2(θ)*(dθ/dt) = (1/1000)(dh/dt)

sec^2(π/3rad)*(0.05rad/s) = (1/1000)*(dh/dt)

Multiplying both sides by 1000

1000*sec^2(π/3rad)*(0.05rad/s) = dh/dt

Simplifying the left side

200 m/s = dh/dt

^answer