An arrow is launched straight up with a velocity of 12.6 m/s. How long before the arrow returned to the ground?

2.6 secs.

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 12.6 m/s

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

Next, we shall determine the time taken to reach the maximum height.

NOTE: the maximum height is reached when the final velocity of the arrow in the air is zero.

Thus, we can obtain the time taken to reach the maximum height as follow:

Initial velocity (u) = 12.6 m/s

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

Final velocity (v) = 0 m/s

Time (t) =?

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

0 = 12.6 – 9.8 × t

0 = 12.6 – 9.8t

Rearrange

9.8t = 12.6

Divide both side by 9.8

t = 12.6/9.8

t = 1.3 secs.

Therefore, the time taken to reach the maximum height (i.e at a point where the final velocity is 0) is 1.3 secs.

Finally, we shall determine the time taken for the arrow to return to the ground as follow:

Time (t) taken to reach the maximum height = 1.3 secs.

Time (T) taken for the arrow to return to the ground =.?

T = 2t

T = 2 × 1.3

T = 2.6 secs.

Therefore, the time taken for the arrow to return to the ground is 2.6 secs.