A rock is thrown straight up with an initial speed of 22 m/s. How long will it be in the air before it returns to the thrower?

The time spent in air by the rock before it returns to the thrower is 4.5 s

Explanation:

Given;

initial velocity of the rock, u = 22 m/s

Neglecting air resistance, the time it takes the rock to go up equals the time it takes it to return to the thrower. This is also called time of flight.

The time to go up is calculated as;

v = u - gt

where;

v is the final velocity at the maximum height

0 = u - gt

gt = u

t = u / g

t = 22 / 9.8

t = 2.245 s

thus time to go up = 2.245 s

and time to return to the thrower = 2.245s

The total time spent in air by the rock before it returns to the thrower

 = 2.245 s + 2.245s

 = 4.5 s