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. how many seconds must the ball have been in the air?

Explanation:

Step1: Find the initial vertical velocity

The initial velocity $v_0 = 13$ m/s and the launch - angle $\theta=45^{\circ}$. The initial vertical velocity $v_{0y}=v_0\sin\theta$. So $v_{0y}=13\sin45^{\circ}=13\times\frac{\sqrt{2}}{2}\approx9.19$ m/s.

Step2: Use the kinematic equation for vertical motion

The kinematic equation for vertical displacement $y - y_0=v_{0y}t-\frac{1}{2}gt^2$. Since the ball starts and ends at ground - level ($y - y_0 = 0$) and $g = 9.8$ m/s², we have $0=v_{0y}t-\frac{1}{2}gt^2$. Factoring out $t$, we get $t(v_{0y}-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the initial time of the kick). The other non - zero solution is $t=\frac{2v_{0y}}{g}$.

Step3: Calculate the time of flight

Substitute $v_{0y}\approx9.19$ m/s and $g = 9.8$ m/s² into $t=\frac{2v_{0y}}{g}$. So $t=\frac{2\times9.19}{9.8}\approx1.88$ s.

Answer:

$1.88$