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Answer :
The rocket returns to the ground after approximately 12.1 seconds. This calculation assumes a symmetrical flight path where the time of ascent equals the time of descent.
To determine the total time the rocket spends in the air, we need to analyze its motion under the influence of gravity. The initial velocity [tex](\(v_0\))[/tex] of the rocket is 195 feet per second, and the acceleration due to gravity [tex](\(g\))[/tex] is approximately 32.2 feet per second squared downward.
First, we find the time to reach the highest point, where the velocity becomes zero. Using the equation [tex]\(v = v_0 - gt\)[/tex]:
[tex]\[ 0 = 195 - 32.2t \][/tex]
[tex]\[ 32.2t = 195 \][/tex]
[tex]\[ t = \frac{195}{32.2} \approx 6.1 \text{ seconds} \][/tex]
This is the time to reach the highest point. Since the time to ascend and descend are equal in the absence of air resistance, the total flight time is twice this duration:
[tex]\[ \text{Total time} = 2 \times 6.1 = 12.2 \text{ seconds} \][/tex]
Rounding to the nearest tenth, the rocket returns to the ground approximately 12.1 seconds after launch.
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