math college algebra

Question

A toy rocket is launched from the top of a building 75 feet tall at an initial velocity of 160 feet per second. The position at time t is given by 𝑠(𝑡) = −16𝑡 2 + 𝑣0𝑡 + 𝑠0 and the velocity at time t is given by 𝑣(𝑡) = −32𝑡 + 𝑣0 where 𝑣0 is the initial velocity and 𝑠0 is the initial height.

Answer 1

in 5 seconds it reaches maximum height

f= -32t +160, t=160/32 = 5 seconds

max height = -16(5)^2 +160(5) + 75 = 475 feet

time to reach the ground is found by setting s = 0 and solving for t

0= -16t2 + 160t +75

16t^2 -160t -75 = 0

t^2 -10t -75/16 = 0

complete the square: take half the x term's coefficient, -5, square it to get 25, add 25 to both sides

t^2 -10t + 25 = 75/16 + 25 factor

(t-5)^2 = 475/16 take the square root of both sides

t-5 = (5/4)sqr19

t =about 5 + or - (5/4)4.36= 5+5.45 = 10,45 seconds to reach the ground (ignore the negative square root)