A soccer player kicks a ball into the air at an angle of 34 degrees above the horizontal . The initial velocity of the ball is 25 m/s How long is the soccer ball in the air ?
vertical:
hf=hi+Vverti*time-4.9t^2
0=0+25sin34*t-4.9t^2
solve for t
To calculate the time the soccer ball is in the air, we can use the following equation of motion:
h = v₀y * t - 0.5 * g * t²,
where
h is the maximum height the ball reaches,
v₀y is the initial vertical velocity component,
g is the acceleration due to gravity (approximately 9.8 m/s²),
and t is the time the ball is in the air.
To find the maximum height, we can use the equation:
v₀y = v₀ * sin(θ),
where
v₀ is the initial velocity of the ball,
and θ is the angle the ball is kicked at.
In this case, v₀ = 25 m/s and θ = 34°.
Therefore,
v₀y = 25 * sin(34°).
Let's calculate v₀y:
v₀y = 25 * sin(34°) = 25 * 0.559 = 13.96 m/s (approximately).
Now, using the equation of motion:
h = (13.96 * t) - (0.5 * 9.8 * t²).
At the maximum height, the ball reaches, h = 0. Therefore,
0 = (13.96 * t) - (0.5 * 9.8 * t²).
Rearranging the equation, we get:
4.9t² - 13.96t = 0.
Factorizing the equation, we get:
t * (4.9t - 13.96) = 0.
So, either t = 0 (initial time) or 4.9t - 13.96 = 0.
Let's solve the second part of the equation:
4.9t - 13.96 = 0.
4.9t = 13.96.
t ≈ 2.85 (approximately).
Therefore, the soccer ball is in the air for approximately 2.85 seconds.