Answer:
a) 252 ft/s
b) 1076 ft
Explanation:
The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:
Y(t) = Y0 + V0 * t + 1/2 * a * t^2
Where
Y(t): altitude at a given time
Y0: initial altitude
V0: initial speed
a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)
We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.
We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was
Y(0) = 89.6 ft
Therefore:
Y0 = 89.6 ft
We also know that the rocket fell to ground 16 seconds later, therefore
Y(16) = 0 ft
So we can write
Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2
V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2
V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t
V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s
In the highest point of flight the rocket will have a speed = 0
The first derivative of the equation of motion is the equation of speed:
V(t) = V0 + a * t
If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.
0 = V0 + a * t
a * t = 0 - V0
t = -V0/a
t = -252/(-32.2) = 7.83 s
Now, we can take this time value andd plug it back into the position equation
Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft
