A soccer ball is kicked with a speed of 22m/s at an angle of 35.0º above the horizontal. If the ball lands at the same level from which it was kicked, how long was it in the air?

2.57 seconds

Explanation:

The motion of the ball on the two axis is;

x (t) = Vo Cos θt

y (t) = h + Vo sin θt - 1/2gt²

Where; h is the initial height from which the ball was thrown.

Vo is the initial speed of the ball, 22 m/s, θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²

We want to find the time t at which y (t) = h

Therefore;

y (t) = h + Vo sin θt - 1/2gt²

Whose solutions are, t = 0, at the beginning of the motion, and

t = 2 Vo sinθ/g

= (2 * 22 * sin 35°) / 9.81

= 2.57 seconds