Atoy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. disregarding any air resistance, what maximum height above the ground will the rocket achieve? the acceleration of gravity is 9.8 m/s 2 .

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

We must consider that it's launched from the ground () and from rest (), with an upwards acceleration that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

And the velocity achieved in part 1:

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 () and its initial velocity is the one achieved in part 1 (), now in free fall, which means with a downwards acceleration . For the data we have it's faster to use the formula , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

Then, to get , we do:

And we substitute the values:

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