Question

A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light. How fast is the shadow of the ball moving along the ground 1 sec​ later? (Assume the ball falls a distance s equals 16 t squared in t​ sec.)

Answer 1

Answer:

$\frac{dx}{dt}=-{187.5} m/sec$

Explanation:

Let x represents the horizontal distance between the shadow and the pole

s= s(t) = 16t^2, distance the ball falls in time t

Then we have $\frac{ds}{dt} = 32t ft/sec$

Also after t=1 sec ball traveled

s= 16× (1)^2 = 16 ft

we need find relation between x and s. So, we do that by using given similar triangles ( first one created by the foot of the pole, top of the pole and shadow of the ball. Second one by the ball, part of the ground that is 30 ft away from the light and the shadow of the ball) By using similarity of triangles we can say

$\frac{x}{x-30}=\frac{50}{50-s}$

⇒ x= 1500/s

Finally, differentiating the above equation we get

$\frac{dx}{dt}=-\frac{1500}{s^2}\frac{ds}{dt}$

$\frac{dx}{dt}=-\frac{1500}{16^2}\times32\times1$

$\frac{dx}{dt}=-{187.5}$