Aball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. the distance s (in feet) of the ball from the ground after t seconds is s equals 64 plus 48 t minus 16 t squared . (a) after how many seconds does the ball strike the ground? (b) after how many seconds will the ball pass the top of the building on its way down?

a) t = 4 secs

b) t = 3 secs

Step-by-step explanation:

h = 64 ft

Initial velocity = 48 ft/s

S = 64 + 48t - 16t^2

a) Set S = 0 and solve for t

0 = 64 + 48t - 16t^2

Divide through by 16

0 = 4 + 3t - t^2

t^2 - 3t - 4 = 0

t^2 - 4t + t - 4 = 0

t(t - 4) + 1(t - 4) = 0

(t + 1) (t - 4) = 0

t + 1 = 0 or t - 4 = 0

t = -1 or t = 4

t = {-1,4}

t = 4 seconds

b) Set S = 64 and solve for t

64 = 64 + 48t - 16t^2

64 - 64 = 48t - 16t^2

0 = 48t - 16t^2

Divide through by 16

0 = 3t - t^2

t^2 - 3t = 0

t(t - 3) = 0

t = 0 or t-3 = 0

t = 0 or t = 3

t = {0,3}

t = 3 secs