SOLUTION: 13. (8 pts) If a projectile is launched from a platform 40 feet high with an initial velocity of 96 feet per second, then the height of the projectile at t seconds is given by s

Algebra ->  Graphs -> SOLUTION: 13. (8 pts) If a projectile is launched from a platform 40 feet high with an initial velocity of 96 feet per second, then the height of the projectile at t seconds is given by s      Log On


   

Question 150020: 13. (8 pts) If a projectile is launched from a platform 40 feet high with an initial velocity of 96 feet per second, then the height of the projectile at t seconds is given by
s(t) = –16t^2 + 96t + 40 feet.
(i) At what time does the projectile attain its maximum height? Show some work.






(ii) What is the maximum height attained by the projectile? Show some work.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 13

In order to find the time it takes to reach the max height, use this formula:

t=%28-b%29%2F%282a%29 Start with the given formula.


From y=-16t%5E2%2B96t%2B40, we can see that a=-16, b=96, and c=40.


t=%28-%2896%29%29%2F%282%28-16%29%29 Plug in a=-16 and b=96.


t=%28-96%29%2F%28-32%29 Multiply 2 and -16 to get -32.


t=3 Divide.


So it takes 3 seconds for the projectile to attain its maximum height.

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ii)

To find the max height, simply plug the time t=3 into the equation y=-16t%5E2%2B96t%2B40



y=-16t%5E2%2B96t%2B40 Start with the given equation.


y=-16%283%29%5E2%2B96%283%29%2B40 Plug in t=3.


y=-16%289%29%2B96%283%29%2B40 Square 3 to get 9.


y=-144%2B96%283%29%2B40 Multiply -16 and 9 to get -144.


y=-144%2B288%2B40 Multiply 96 and 3 to get 288.


y=184 Combine like terms.


So the max height is 184 feet.