An apple is falling from an apple tree next to a lamppost 24 feet tall. When the

Question

An apple is falling from an apple tree next to a lamppost 24 feet tall. When the apple hits the ground, it will be 4 feet away from the base of the lamppost. The height of the apple is given by H(t) = 35-12t^2 , where t is seconds after the apple begins to fall. At what rate is the apple's shadow moving along the ground when it is 20 feet above the ground? (Hint: First find the time when the apple is 20 feet above ground, and use that answer to find dh/dt at the moment it is 20 feet above ground)

Please show every step.

Explanation / Answer

When the apple is at a height of 20feet above th ground, 20=35-12t^2 solving gives t=sqrt(5)/2
and at the same instant the apple is 4 ft below tha lamp post(taking only in vertical direction) and the apple is horizontally 4ft away from the lamp post(mentioned in the question). So at that instant the rate of hieght of the apple is equal to the rate of shadow .(dH/dt at t=srqt(5)/2 is same as dS/dt at the same time, where S denotes the length of shadow).
dH/dt=-24t
dS/dt (at that time)=dH/dt (at that time t) = -24 sqrt(5)/2 = -12sqrt(5)

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