To find how long the ball was in the air, you can use the kinematic equations of motion.
Step 1: Split the initial velocity into horizontal and vertical components.
The horizontal component of the velocity can be found using the equation:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the initial speed (22 m/s), and θ is the angle (35.0º) above the horizontal.
Vx = 22 m/s * cos(35.0º)
Vx ≈ 17.94 m/s
The vertical component of the velocity can be found using the equation:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the initial speed (22 m/s), and θ is the angle (35.0º) above the horizontal.
Vy = 22 m/s * sin(35.0º)
Vy ≈ 12.63 m/s
Step 2: Use the vertical component of the velocity to find the time of flight.
Since the ball lands at the same level from which it was kicked, it means the vertical displacement is zero.
The equation that relates the vertical displacement (y), initial vertical velocity (Vy), time of flight (t), and acceleration due to gravity (g ≈ 9.8 m/s^2) is:
y = Vy * t + (1/2) * g * t^2
Substituting the values, we have:
0 = 12.63 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation, we get:
4.9 * t^2 + 12.63 * t = 0
Step 3: Solve the quadratic equation to find the time of flight.
Factoring out t, we have:
t * (4.9 * t + 12.63) = 0
So, either t = 0 or (4.9 * t + 12.63) = 0
Since time cannot be negative, we can calculate t by solving the second equation:
4.9 * t + 12.63 = 0
Subtracting 12.63 from both sides:
4.9 * t = -12.63
Dividing both sides by 4.9:
t ≈ -2.58 seconds
Since time cannot be negative in this context, we discard this solution.
Therefore, the ball was in the air for approximately 0 seconds.