A projectile is fired upward 200 ft from an observer. The height in feet of the projectile after t seconds is given by s=-16t2 + 180t. What is the rate of change of the angle of elevation of the observer to the projectile after 7 seconds?

I don't understand how to get there!

if θ is the angle of elevation,

y = -16t^2 + 180t
tanθ = y/200

sec^2θ dθ/dt = 1/200 dy/dt = 1/200 (-32t + 180)
= -32/200 (t-5)

dθ/dt = -4/25 (t-5) cos^2θ
= -4/25 (t-5) (200^2/(200^2 + y^2)
y(7) = -16*49 + 180*7 = 476

at t=7,
dθ/dt = -4/25 * 2 * 40000/266576 = -.048

Seems kinda small, so you better check the details.

To find the rate of change of the angle of elevation, we need to first determine the equation that relates the angle of elevation to time.

The angle of elevation (θ) can be defined as the angle between the horizontal line (ground) and the line of sight from the observer to the projectile. In a right triangle, the opposite side to θ is the height of the projectile and the adjacent side is the distance between the observer and the projectile.

In this case, we already know that the height of the projectile is given by s = -16t^2 + 180t, so we just need to determine the distance between the observer and the projectile.

Since the projectile was fired upward from the observer, we end up with a vertical motion only. Therefore, the distance (d) between the observer and the projectile is simply the horizontal distance covered by the projectile.

To find this, we need to determine the horizontal component of the projectile’s velocity. Initially, the projectile has no horizontal velocity, so the only horizontal velocity it has is the result of its motion caused by gravity throughout the flight. We can consider this component as constant, as there are no external horizontal forces acting on the projectile.

Using the equation of motion in the horizontal direction, we have:

d = v * t,

where d is the horizontal distance, v is the horizontal component of the projectile's velocity, and t is the time.

To find v, we use the fact that the total time of flight is the same for the vertical and horizontal components of motion.

Since the maximum height is reached when the vertical velocity becomes zero, we can use the equation for vertical velocity, v = u + at, where u is the initial vertical velocity (fired upward, so u = 0) and a is the vertical acceleration (-32 ft/s^2, considering gravity).

At the maximum height, the vertical velocity is zero, so we have:

0 = 0 + (-32)t_max
=> t_max = 0.

This means that the total time of flight is zero, so the horizontal distance covered is also zero. Therefore, d = 0.

Now that we know the distance between the observer and the projectile is zero, we can conclude that the angle of elevation is always 90 degrees.

Therefore, the rate of change of the angle of elevation is zero at all times, including 7 seconds.