Question

A light shines from the top of a pole 30 ft high. A ball is dropped from the same height from a point 20 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

After 1 seconds, the speed of the ball's shadow along the ground is equal to 150 ft/s.

How to calculate the speed of the ball's shadow?

Assuming the ball falls a distance given by this expression:

S = 16t²

Also, the height (h) of shadow of the ball 1 sec​ later is given by:

h = 30 - S

Next, we would derive an expression for the ball's shadow:

OX/30 = (20 + XQ)/30 = XQ/h

30XQ = 20h + hXQ

20h = 30XQ - hXQ

20h = (30 - h)XQ

XQ = 20h/(30 - h)

Substituting the value of h, we have:

XQ = 20(30 - S)/(30 - 30 - S)

Substituting the value of S, we have:

XQ = 20(30 - 16t²)/(30 - 30 - 16t²)

XQ = 20(30 - 16t²)/16t²

XQ = (600 - 320t²)/16t²

XQ = 600/16t² - 320t²/16t²

XQ = 600/16t² - 20

Differentiating wrt t, we have:

d(XQ)/dt = 600/16t² - 20

d(XQ)/dt = 600 (-64t/(16t²)²)

d(XQ)/dt = 600 (-64t/256t⁴)

d(XQ)/dt = 600 (-0.5/2t³)

d(XQ)/dt = -300/2t³

At t = 1, we have:

Speed = 300/2(1)

Speed = 150 ft/s.

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