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Sandra shot a rocket so that it moved with an initial velocity of 9.81 m/s straight upward. The rocket leaves and returns to ground level. What is the total time the rocket was in the air before it strikes the ground? (Ignore air resistance.

Sandra shot a rocket so that it moved with an initial velocity of 9.81 m/s straight-example-1
User Jbccollins
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1 Answer

2 votes

Answer:

2.00 seconds.

Step-by-step explanation:

The following data were obtained from the question:

Initial velocity (u) = 9.81 m/s

Total time in air (T) =..?

Next, we shall determine the time taken for the rocket to reach it's maximum height. This can be obtained as follow:

Note: At maximum height, the final velocity is zero.

Initial velocity (u) = 9.81 m/s

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time to reach the maximum height (t) =.?

v = u – gt (since the rocket is going against gravity)

0 = 9.81 – 9.8t

Rearrange

9.8t = 9.81

Divide both side by 9.8

t = 9.81/9.8

t = 1.00 s

Therefore the time taken to reach the maximum height is 1.00 second.

Finally, we shall determine the total time spent by the rocket in the air as follow:

Time to reach the maximum height (t) = 1.00 s

Total time in air (T) =..?

T = 2t

T = 2 × 1.00

T = 2.00 s

Therefore, the total time spent by the rocket in the air is 2.00 seconds.

User Mahmoud Aladdin
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