The path of a rocket fired during a fireworks display is given by the equation s(t)=64t2,( t do the 2nd power) where t is the time in seconds and s in the height in feet. What is the maximum height in feet the rocket will reach? In how many seconds will the rocket hit the ground?

Question

I want the solution with all the appropriate work shown how to solve this problem.

Answer 1

Check your equation.
If it is 64 t^2 it goes up forever.
However it is not.
It is something of form
s(t) = Vo t - (1/2) g t^2
where g, gravity acceleration near earth, is 32 ft/s^2 or 9.8 m/s^2

Answer 2

The path of a rocket fired during a fireworks display is given by the equation s(t)=64t2-16t2,( t do the 2nd power) where t is the time in seconds and s in the height in feet. What is the maximum height in feet the rocket will reach? In how many seconds will the rocket hit the ground?

Answer 3

Are you sure it is not
s = 64 t - 16 t^2 ???
I will assume that to be the case.
That is a parabola
find the vertex
16 t^2 - 64 t = -s

t^2 - 4 t = -s/16

t^2 - 4 t + 4 = -s/16 + 4

(t-2)^2 = -(1/16)(s-64)
That is a parabola opening down (sheds water) with vertex at t = 2, s =64
so
top at t=2 and s = 64
now you could say it will spend the same time falling as rising (symmetry) so will hit the ground when t=4 seconds.
however look when s = 0
(t-2)^2 = -(-4)
(t-2)= +2 or -2
when t = 0 and when t 4
so hits when t = 4