Question

4 numeri 2 point 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. answer

Explanation:

Step1: Recall projectile range formula

The range \( R \) of a projectile launched from ground level (initial and final height \( y = 0 \)) with initial velocity \( v_0 \) at angle \( \theta \) is given by \( R=\frac{v_0^2\sin(2\theta)}{g} \), where \( g = 9.8\space m/s^2 \) is the acceleration due to gravity.

Step2: Substitute values

Given \( v_0 = 13\space m/s \), \( \theta = 45^\circ \). First, calculate \( \sin(2\theta) \). For \( \theta = 45^\circ \), \( 2\theta = 90^\circ \), and \( \sin(90^\circ)=1 \). Then substitute into the formula: \( R=\frac{(13)^2\times1}{9.8} \).

Step3: Calculate numerator and divide

\( (13)^2 = 169 \). Then \( R=\frac{169}{9.8}\approx17.24\space m \) (or more precisely, \( \frac{169}{9.8}=\frac{1690}{98}=\frac{845}{49}\approx17.24 \)).

Answer:

\(\approx 17.24\) (or more accurately, \(\frac{845}{49}\) or approximately \(17.2\) - \(17.3\) meters)