Question

a football is kicked from ground level and has an initial velocity of 13 m/s directed at an angle of 45 degrees from the horizontal. the football travels in an arc through the air and eventually lands on the ground again. what must be the range of the football in meters? remember, the range is the distance traveled along the ground in the x direction.

Explanation:

Step1: Identify the range formula

The range formula for projectile motion is $R=\frac{v_{0}^{2}\sin2\theta}{g}$, where $v_{0}$ is the initial velocity, $\theta$ is the launch - angle, and $g = 9.8\ m/s^{2}$ is the acceleration due to gravity.

Step2: Substitute the given values

Given $v_{0}=13\ m/s$ and $\theta = 45^{\circ}$, then $\sin2\theta=\sin(2\times45^{\circ})=\sin90^{\circ}=1$.
Substitute into the formula: $R=\frac{(13)^{2}\times1}{9.8}$.

Step3: Calculate the range

$R=\frac{169}{9.8}\approx17.24\ m$.

Answer:

$17.24$