Part A
Suppose a rocket is being launched into outer space. On its way up, several different forces act on the rocket. One of these forces, thrust, is
applied by the rocket's engines.
The rocket's engines apply a constant upward force to a rocket on its way to outer space. In the absence of other forces, the upward force on the
rocket would cause it to accelerate up at a constant rate of a meters/second.
Still ignoring any other forces, the height of the rocket can be expressed as a function of the time elapsed in seconds after the launch, t. This
function, ft), is equal to half the product of the rocket's acceleration and the square of the time after launch. Write the equation for ft).

[tex]\displaystyle \mathrm{The \ function \ that \ gives \ the \ height \ of \ the rocket \ is \ }\underline{f(t) = \frac{1}{2} \cdot a \cdot t^2}[/tex]

Reasons:

The acceleration of the rocket = a

According to Newton's Law of motion, the height reached by the rocket with time can be found by the formula;

[tex]\displaystyle f(t) = u \cdot t - \frac{1}{2} \cdot a \cdot t^2[/tex]

Where;

u = The initial velocity of the rocket = 0

Which gives;

[tex]\displaystyle f(t) = 0 \times t - \frac{1}{2} \cdot a \cdot t^2 = \mathbf{ \frac{1}{2} \cdot a \cdot t^2}[/tex]

[tex]\displaystyle \underline{The \ height \ of \ the \ rocket, \ f(t) = \frac{1}{2} \cdot a \cdot t^2}[/tex]

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https://brainly.com/question/3069698