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------------------------------------------------ What would be the height if your rocket were in the air for 5.2 seconds?

Use the formula:
[tex]\[ h = \frac{1}{2} a t^2 + v_0 t + h_0 \][/tex]

A. 30.6
B. 34.9
C. 39.1
D. None of the above

Let's determine the height of the rocket after 5.2 seconds using the equation:

[tex]\[ h = \frac{1}{2} a t^2 + v_0 t + h_0 \][/tex]

Here are the given values:

- [tex]\( a \)[/tex] (acceleration) = 0, meaning there are no external forces acting on the rocket vertically.
- [tex]\( t \)[/tex] (time) = 5.2 seconds
- [tex]\( v_0 \)[/tex] (initial velocity) = 0, implying the rocket started from rest.
- [tex]\( h_0 \)[/tex] (initial height) = 0, indicating the rocket launched from ground level.

Substitute these values into the equation:

[tex]\[ h = \frac{1}{2} \times 0 \times (5.2)^2 + 0 \times 5.2 + 0 \][/tex]

Simplifying it:

[tex]\[ h = 0 + 0 + 0 \][/tex]

[tex]\[ h = 0 \][/tex]

So, the height of the rocket after 5.2 seconds is 0.

Among the options provided (30.6, 34.9, 39.1, None), the correct option is None.