Question 1203587
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Part (a)


s = -16t^2 + v0*t
s = -16t^2 + 144t
288 = -16t^2 + 144t
0 = -16t^2 + 144t - 288
-16t^2 + 144t - 288 = 0


We can then factor
-16t^2 + 144t - 288 = 0
-16(t^2 - 9t + 18) = 0
-16(t - 3)(t - 6) = 0
t-3 = 0 or t-6 = 0
t = 3 or t = 6



Or we can use the quadratic formula
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-144+-sqrt((144)^2-4(-16)(-288)))/(2(-16))}}} Plugged in a = -16, b = 144, c = -288


{{{t = (-144+-sqrt(20736 - 18432))/(-32)}}}


{{{t = (-144+-sqrt(2304))/(-32)}}}


{{{t = (-144+-  48)/(-32)}}}


{{{t = (-144+48)/(-32)}}} or {{{t = (-144-48)/(-32)}}}


{{{t = (-96)/(-32)}}} or  {{{t = (-192)/(-32)}}}


{{{t = 3}}} or  {{{t = 6}}}



Either way, the two answers for part (a) are t = 3 and t = 6


The projectile reaches 288 ft at t = 3 seconds when going upward. 
Upon its downward trajectory, it gets back to a height 288 ft at t = 6 seconds.


The projectile is above 288 ft for the interval 3 < t < 6. Otherwise, the projectile is at 288 ft or lower.


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Part (b)


s = -16t^2 + 144t
-16t^2 + 144t = s
-16t^2 + 144t = 0
-16t(t - 9) = 0
-16t = 0 or t-9 = 0
t = 0/(-16) or t = 9
t = 0 or t = 9


The projectile starts on the ground, so it makes sense that t = 0 is one solution.


The other solution is t = 9 which is when the rocket returns to the ground again. 
The flight time is 9 seconds.
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