a football player kicks a football with a speed of 21.0m/s at 37.0 degrees to the horizontal how long is the football in the air
To find the total time the football is in the air, we can break the initial velocity into its horizontal and vertical components.
Horizontal component: $v_{0x} = v_0 \cdot cos(\theta)$
$v_{0x} = 21.0 \cdot cos(37.0)$
$v_{0x} = 21.0 \cdot 0.7986$
$v_{0x} = 16.77 \, m/s$
Vertical component: $v_{0y} = v_0 \cdot sin(\theta)$
$v_{0y} = 21.0 \cdot sin(37.0)$
$v_{0y} = 21.0 \cdot 0.6018$
$v_{0y} = 12.64 \, m/s$
We can ignore the vertical component when calculating time in the air, as it will reach a height of 0 at the same time it returns to the ground.
Using the horizontal component of velocity, we can find the time the ball is in the air using the equation:
$R = v_{0x} \cdot t$
Where R is the range of the projectile and t is the time in the air.
The range of the projectile is the horizontal distance the football travels. This can be found using the equation:
$R = \dfrac{v^2_{0x} \cdot sin(2\theta)}{g}$
Where g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values, we get:
$R = \dfrac{(16.77)^2 \cdot sin(74)}{9.8}$
$R = \dfrac{281.5 \cdot 0.9624}{9.8}$
$R = \dfrac{271.0424}{9.8}$
$R = 27.66 \, m$
Now that we know the range, we can find the time the ball is in the air using:
$27.66 = 16.77 \cdot t$
$t = \dfrac{27.66}{16.77}$
$t = 1.65 \, s$
Therefore, the football is in the air for approximately 1.65 seconds.