Question

The height s of a ball (in feet) thrown with an initial velocity of 64 feet per second from an initial height of 7 feet is given as a function of the time t (in seconds) by
s(t)=-16t^2+64t+7
Graph s(t) and answer;
a)Determine the time at which height is maximum
b)What is the maximum height.

Answer 1

take derivative to this function that is the t is the time

Answer 2

I am suppose to be entering this in ti83 and finding the answer...

Answer 3

The solutions of the quadratic equation ax2 + bx + c = 0 are
x = when a ≠ 0 and b2 º 4ac ≥ 0.
You can read this formula as “x equals the opposite of b, plus or
minus the square root of b squared minus 4ac, all divided by 2a.”
ºb ± b2º 4 ac 2a

Answer 4

\[s^'(t)=0\] speed zero at max height

Answer 5

\[s^'(t)=-(2)16t+64=0\]

Answer 6

b)use the t above in \[s()=-16( )^2+64()+7\]
empty space is yur time above

Answer 7

Thats Right But You Look My Answer thats Really Right

Answer 8

I am really confused, but here is an example. So is this close to what you guys are saying.

Answer 9

its not confusing the question is basically asking for the Turning point (bvertex can you find that)

Answer 10

did you do calculus derivatives and its application

Answer 11

Jonask : Good Qouestion

Answer 12

no, I don't think so

Answer 13

ok so can you find a vertex of a parabola

Answer 14

y=-16x^2+64x+7

Answer 15

@Jonask so is it 2

Answer 16

I think I figured this out. So I solved the quad. formula it came out to 2 and I plug the 2 in -16(2)^2+64(2)+7=71, which gives me the answer to the second part.

Answer 17

yes

Answer 18

thanks for the help