After being thrown from the top of a tall building, a projectile follows the pat

ID: 2874619 • Letter: A

Question

After being thrown from the top of a tall building, a

projectile follows the path (x, y) = (60t, 784 16t^2) in

which x and y are in feet and t is in seconds. The Sun is

directly overhead, so that the projectile casts a moving

shadow on the ground beneath it, as shown in the figure.

What is the altitude of the projectile when t = 2? What is the altitude

of the projectile a little later, when t = 2 + k? How much altitude is lost during this

k-second interval? At what rate is the projectile losing altitude during this interval?

1. Altitude is the y value . Hence y ( 2 ) = 784 - 16 x4 = 784 - 64 720 feet

2. y ( 2 + k ) = 784 - 16( 2 + k ) 2 = 784 - 64 - 64k - 16 k2

= 720 - 64 k - 16 k2

3. altitude lost in k seconds = - 64 k - 16 k2 feets

4. rate at which the projectile losing altitude is dy /dk = - 64 - 32k

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