Another Related Rates question

[fallen_6666]

A light is at the top of a pole 20 m high. From a point on the ground 8m from the base of the pole a ball is thrown upwards with an inital velocity of 13 m/s. At what rate is the shadow of the ball moving along the ground 1 second later.

[woelen]

In order to solve this problem, a few questions arise

1) Is there any gravitational pull and hence, is there a gravitational acceleration towards to ground? What is the value of this pull (9.8 m/s2)?

2) The speed of light is considered infinite?

3) The size of the ball can be neglected, i.e. it can be considered a point mass?

 

Given these assumptions, the problem is fairly simple.

 

Compute where the ball is after one second:

 

dv/dt = -9.8 m/s2 ==> v(t) = v(0) - 9.8t = 13 - 9.8t (m/s)

dh/dt = v ==> h(t) = h(0) + v(0)*t - 4.9*t*t = 0 + 13*t - 4.9*t*t (m)

 

Plug in t = 1. This gives you the height and the vertical velocity.

 

Now derive an expression for the distance of the shadow from the pole as function of h. This involves goniometry and is left for the OP as an exercise. You get some expression as function of h, lets call this d(h).

 

The velocity at which the shadow is moving along the floor can be written as

 

dd/dt = d'(h) * dh/dt = d'(h)*v, where d'(h) is the derivative of the function d with respect to h.

 

Plug in the value of h at d'(h) and take the value of v, both at t = 1 second. That gives you the result.