A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. How long was the ball in the air?
To find out how long the ball was in the air, we need to use the equations of motion for projectile motion. In this case, we're dealing with a projectile being launched at an angle.
The horizontal motion and vertical motion of the projectile are independent of each other. So, let's first break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity can be found using the formula: Vx = V * cos(θ)
Vx = 26 m/s * cos(30°)
= 26 m/s * 0.866
≈ 22.44 m/s (rounded to two decimal places)
Here, θ represents the launch angle of 30 degrees.
The vertical component of the initial velocity can be found using the formula: Vy = V * sin(θ)
Vy = 26 m/s * sin(30°)
= 26 m/s * 0.5
= 13 m/s
Now, using the vertical motion of the projectile, we can find the time of flight. The equation for the vertical displacement, or height, of a projectile at any time (t) can be expressed as:
y = Vyt + (1/2) * g * t^2
Since the projectile started from the ground, the initial vertical displacement (y) is 0. The final vertical displacement (y) when the ball hits the ground is again 0.
So, we have:
0 = 13 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Rearranging the equation, we get:
(1/2) * (-9.8 m/s^2) * t^2 + 13 m/s * t = 0
Simplifying further, we have a quadratic equation:
-4.9 * t^2 + 13 * t = 0
Now, we can solve for t. We can factor out t:
t * (-4.9 * t + 13) = 0
So, either t = 0 (which is not possible since it represents the initial time), or -4.9 * t + 13 = 0
Solving the equation -4.9 * t + 13 = 0 for t, we get:
-4.9 * t = -13
Dividing both sides by -4.9, we get:
t = -13 / -4.9
≈ 2.65 seconds (rounded to two decimal places)
Therefore, the ball was in the air for approximately 2.65 seconds.