Imagine that you throw a rock into the air. When you release the rock, it is moving at 19.4 m/s. The rock flies straight up into the air and returns back to your hand. For how long was the rock in the air? Note: acceleration due to gravity is -9.81 m/s/s.

4 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 19.4 m/s.

Acceleration due to gravity (g) = –9.81 m/s²

Total time (T) =.?

Next, we shall determine the time take by the rock to get to the maximum height. This can be obtained as follow:

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

Initial velocity (u) = 19.4 m/s.

Acceleration due to gravity (g) = –9.81 m/s²

Final velocity (v) = 0 m/s

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

v = u + gt

0 = 19.4 + (–9.81 × t)

0 = 19.4 – 9.81t

Collect like terms

0 – 19.4 = –9.81t

–19.4 = –9.81t

Divide both side by –9.81

t = –19.4 / –9.81

t ≈ 2 s

Thus, it took approximately 2 s for the rock to reach its maximum height.

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

Time taken to reach the maximum height (t) = 2 s

Total time spent in air (T) =?

The total time (T) spent by the rock in the air will be two times the time taken (t) to reach the maximum height i.e

T = 2t

t = 2 s

T = 2 × 2

T = 4 s

Therefore, the rock spent approximately 4 s in the air.

v=ir = 3x15 =45 v