To find the time the ball is in the air, we can analyze the vertical motion of the ball. We can break down the initial velocity into its x and y components.
Given:
Initial velocity (v) = 32 m/s
Launch angle (θ) = 17.9°
The initial vertical velocity (Vy) can be found using trigonometry:
Vy = v * sin(θ)
Vy = 32 * sin(17.9°)
Vy ≈ 9.0 m/s
The time of flight (t) can be determined using the vertical motion equation:
y = Vy * t + (1/2) * g * t^2
Since the ball lands at the same height where it was kicked from, the change in vertical position (y) is zero. Therefore, we can solve the equation for t:
0 = Vy * t + (1/2) * g * t^2
Where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Substituting the known values:
0 = 9.0 * t + (1/2) * 9.8 * t^2
Simplifying the equation:
4.9 * t^2 + 9.0 * t = 0
Since this is a quadratic equation, we can factor it:
t * (4.9 * t + 9.0) = 0
The solutions are:
t = 0 (initial time, not relevant)
4.9 * t + 9.0 = 0
Solving for t:
4.9 * t = -9.0
t = -9.0 / 4.9
Since time cannot be negative in this context, we discard the negative solution. Therefore, the time the ball is in the air is approximately:
t ≈ 1.84 seconds
To find the magnitude of the displacement (distance) of the ball from the kick to landing, we can analyze the horizontal motion of the ball.
The horizontal velocity (Vx) can be found using trigonometry:
Vx = v * cos(θ)
Vx = 32 * cos(17.9°)
Vx ≈ 30.1 m/s
The displacement (distance) can be determined using the horizontal motion equation:
x = Vx * t
Substituting the known values:
x = 30.1 * 1.84
Calculating the displacement:
x ≈ 55.4 meters
Therefore, the magnitude of the displacement of the ball from kick to landing is approximately 55.4 meters.