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