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?
I want the solution with all the appropriate work shown how to solve this problem.
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
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?
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
To find the maximum height reached by the rocket, we need to determine the vertex of the parabolic function represented by the equation s(t) = 64t^2.
The vertex of a parabola in the form y = ax^2 + bx + c can be found using the formula x = -b / (2a), where x represents the time and y represents the height in this case.
Comparing the equation s(t) = 64t^2 to the general form y = ax^2, we can see that a = 64, b = 0, and c = 0. Substituting these values into the vertex formula, we have:
t = -0 / (2 * 64)
t = 0
Therefore, the time at which the rocket reaches its maximum height is t = 0 seconds.
To find the height at this time, we substitute t = 0 into the equation:
s(0) = 64 * 0^2
s(0) = 0
Hence, the maximum height reached by the rocket is 0 feet.
Now, to determine the time it takes for the rocket to hit the ground, we set s(t) = 0 and solve for t:
0 = 64t^2
Dividing both sides of the equation by 64, we have:
t^2 = 0
Taking the square root of both sides, we get:
t = 0
Therefore, the rocket hits the ground at t = 0 seconds.
So, the maximum height reached by the rocket is 0 feet, and it hits the ground at t = 0 seconds, according to the given equation.