Question

Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 48 ft/s from the top of a building 316 ft high, then its height h above the ground t seconds later will be h = 316 + 48t − 16t2. During what time interval (in seconds) will the ball be at least 28 ft above the ground? (Enter your answer using interval notation.)

Answer 1

Answer: t∈[0.6] 0 c≤t≤6 c

Explanation:

h=316+48t-16t²

-16t²+48t+316≥28

-16t²+48t+316-28≥28-28

-16t²+48t+288≥0

Divide both parts of the equation by 16 :

-t²+3t+18≥0

Multiply both parts of the equation by -1 (in this case, change the sign of the inequality to the opposite)

-(-1)t²+(-1)3t+(-1)18≤0

t²-3t-18≤0

t²-6t+3t-18≤0

t(x-6)+3(t-6)≤0

(t-6)(t+3)≤0

Find the zeros of the function:

t-6=0

t-6+6=0+6

t=6

t+3=0

t+3-3=0-3

t=-3

⇒ - ∞__+__-3__-__6__+__+∞

⇒ t∈[-3;6]

t≥0

⇒ t∈[0,6]