A ball is thrown downward with a speed of 8 ft/s from the top of a 68 ft tall building. After t seconds, its height above the ground is given by s(t) = −16t2 − 8t + 68.

Question

A ball is thrown downward with a speed of 8 ft/s from the top of a 68 ft tall building. After t seconds, its height above the ground is given by s(t) = −16t² − 8t + 68.

(a)Determine how long (in s) it takes for the ball to hit the ground.

(b)Determine the velocity (in ft/s) of the ball when it hits the ground.

Answer 1

s(t) = -16t^2 - 8t + 68

it hits the ground when s(t)=0

-16t^2 -8t +68 = 0

4t^2 +2t - 17 = 0

t = -2/8 + (1/8)sqr(4+16(17)) =-.25 + (1/8)sqr276=about-.25 +16.5/8

= about 1.8125

t = about 1.8 seconds to hit the ground

-16(1.8125)^2 -8(1.8125) + 68 = 0

v(t) = s'(t) = -32t-8

-32(1.8125)- 8 = -66 feet per second velocity when it hits the ground