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# A water rocket, launched from the ground, rises verticallywith acceleration of 30 /s2 for 1 s when it runs out of fuel. Disregarding air resistance, how high will the rocket rise?Comments

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A water rocket, launched from the ground, rises verticallywith acceleration of 30 /s2 for 1 s when it runs out of fuel. Disregarding air resistance, how high will the rocket rise? Comments

2020-10-26

1. The distance traveled during the first second (positiveacceleration): $$\displaystyle\triangle{y}={v}_{{{0}}}{t}+{\frac{{{1}}}{{{2}}}}{a}{t}^{{{2}}}={0}+{\frac{{{1}}}{{{2}}}}{\left({30}\right)}{\left({1}^{{{2}}}\right)}={15}{m}$$
The speed after this first second: $$\displaystyle\triangle{v}^{{{2}}}={2}{a}\triangle{y}$$ $$\displaystyle{v}^{{{2}}}={{v}_{{{0}}}^{{{2}}}}+{2}{a}\triangle{y}={0}+{2}{\left({30}\right)}{\left({15}\right)}$$ v=30 m/s
The additional distance traveled (acceleration of g in thenegative direction). At the peak of the flight, the velocityis zero:
$$\displaystyle\triangle{v}^{{{2}}}={2}{a}\triangle{y}$$
$$\triangle y=\frac{\triangle v^{2}}{2a}=\frac{0-v^{2}}{2(-g)}=\frac{v^{2}}{2g}=\frac{30^{2}}{2(9.8)}=45.92$$
Other equation(s) of motion could have also been used to findthis distance.
Total distance traveled: y=15+45.92=60.92m 2. $$\displaystyle{d}={v}_{{{0}}}{t}+{\frac{{{1}}}{{{2}{a}{t}^{{{2}}}}}}$$ the initial velocity is zero. I am going to assume that theacceleration is the average acceleration of the rocket, that is thevertical acceleration minus the acceleration of gravity. I am also assumeing that the acceleration is in meters/secondsquared. this means that:
$$\displaystyle{d}={0}+{\frac{{{1}}}{{{2}}}}{\left({30}\right)}{\left({1}\right)}={15}{m}$$