PreCalculus
posted by Ashley .
Hello,
A projectile is launched straight up from the ground with an initial velocity of 320 ft/sec.
a. When will the projectiles height above the ground be 1024 ft?
For this I got 16 seconds
b. When will the height of the projectile be at least 24 ft?
I am not sure how to figure this one out.
c. When will the object strike the ground?
I got 20 seconds
d. What is the maximum height of the projectile?
I got 1600 feet.
So, I am still unsure about b and I wanted to know if my other answers seem correct
Thanks

c) and d) are correct
for a) and the others I assume you used Calculus and the equation
distance = 16t^2 + 320t to find your answers
you must have set 1024 = 16t^2 + 320t
after rearranging and dividing by 16 this would give you
t^2  20t + 64 = 0
(t16)(t4) = 0
so t = 16 or t = 4
you had the t=16, but failed to show the t=4 case.
It reached a height of 1024 twice, at 4 seconds after liftoff on its way up, and then 16 seconds later on its way down
for b)
set 24 = 16t^2 + 320t
16t^2  320t + 24 = 0
2t^2  40t + 3 = 0
using the quadratic formula
t = (40 ± √(16004(2)(3))/4
= 19.92 or .075
so between the times of .075 and 19.92 seconds the object is at least 24 feet high. 
My mistake. b should have asked at least 1024 ft. So does that mean the answer would be anytime in between 4 and 16 seconds?
b. When will the height of the projectile be at least 1024 ft?
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