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

Answer 1

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.