Question

The altitude in feet of a projectile t seconds after the launch is given by S(t)=-16t^2+144t+32. Find the maximum height reached by the projectile.

Answer 1

First differentiate S with respect to t. Can you do this step?

Answer 2

s speed t is time

Answer 3

In this question S is defined as altitude in feet

Answer 4

So we find S' by a basic application of the power rule, resulting in S'(t)=-32t+144. We then find a critical point of the function by setting S'(t)=0, so 32t=144, and t=4.5. We then find the maximum height, or critical value, by substituting this value for t back into S(t), which I will leave to you.

Answer 5

why is it -32t not -16

Answer 6

The derivative of \(ax^n\) is \(nax^{n-1}\). So in this case, \(2\cdot-16t\).

Answer 7

gotcha

Answer 8

its easy... find the line of symmetry

t = -b/2a t = -144/(2 x 16)

max height when t = 4.5 seconds

alternatively

differentiate and solve for t

this is the stationary point

-32t + 144 = 0

t = 4.5

same result

Answer 9

oops its -144/(2 x -16) gives t = 4.5

Answer 10

s=79.11

Answer 11

thats how high it will reach

Answer 12

I got 356 feet

Answer 13

\[
-16(4.5^2)+144(4.5)+32=356
\]

Answer 14

so dont put 4.5 in with the s in front s(t) thats how i came up with 79.11 divided 356 by 4.5

Answer 15

so the right answer is 356 feet

Answer 16

The correct answer is 356.

Answer 17

thank you cambell st nbouscal and kropot72