A light pole of height 15 meter has a lamp hanging at its top . from a distance 12 meter from the base of the pole , a ball is thrown vertically upwards with velocity 5m/sec . find the rate at which the shadow of the ball on the earth moves away from the base of the pole after 0.5 second .

Answer 1

velocity ball=5-9.8t^2 upward....

draw the figure, if x is the distance from the pole to the shadown, one has two similar triangles

15/x=h/(x-12) where h is the height of the ball.

a) 15x-180=xh
but h=5t-1/2 9.8 t^2
dh/dt=5-9.8t

starting with the equation a)
15 dx/dt=hdx/dt + x dh/dt
or dx/dt (15-h)=x dh/dt
well, we know h as a function of t
and we know x :(x=180/(15-h):
dx/dt= x/(15-h) dh/dt
= 180/(15-h)^2 *(5-9.8t)
then put in for h (5t-1/2 9.8 t^2)
and you have it.

Answer 2

Thank you :)