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A stomp rocket is a toy that is launched into the air from the ground by a sudden burst of pressure exerted by stomping on a pedal. If the rocket is launched at 24 feet per second, how long will it be in the air?

Answer :

The time period for which the rocket is in the air will be 1.408 sec. The equation of motion is used to solve the problem.

What is the time period?

The total time period is the sum of the time required to move upwards and the time required to move downwards.

The height attained by the rocket is found as;

[tex]\rm v^2 = u^2 -2gh \\\\ \rm u^2= 2gh \\\\ h=\frac{u^2}{2g} \\\\ h=\frac{(24\times 0.3048)^2}{2\times 9.81 } \\\\ h=2.72 \ m[/tex]

The time for the rocket in the air is found as;

[tex]\rm v= u+gt \\\\ \rm t = \frac{v-u}{g} \\\\ t = \frac{0-24 \times 0.3048 }{-9.81} \\\\ t= 0.704\sec\\\\ T=2\times t \\\\ T= 2\times 0.704 \\\\ T=1.408 \ sec[/tex]

Hence, the time period for which the rocket is in the air will be 1.408 sec.

To learn more about the time period, refer to the link;

https://brainly.com/question/569003

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