Question

7 numeric 1 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. how many seconds must the ball have been in the air? answer previous next

Explanation:

Step1: Find vertical component of velocity

The initial velocity \( v_0 = 13 \, \text{m/s} \) at an angle \( \theta = 45^\circ \). The vertical component \( v_{0y} = v_0 \sin\theta \). So \( v_{0y} = 13 \sin(45^\circ) = 13 \times \frac{\sqrt{2}}{2} \approx 9.192 \, \text{m/s} \).

Step2: Use kinematic equation for vertical motion

The vertical displacement \( y - y_0 = 0 \) (starts and lands at ground level). The kinematic equation is \( y - y_0 = v_{0y}t - \frac{1}{2}gt^2 \), where \( g = 9.8 \, \text{m/s}^2 \). Substituting \( y - y_0 = 0 \), we get \( 0 = 9.192t - 4.9t^2 \). Factoring out \( t \): \( t(9.192 - 4.9t) = 0 \). One solution is \( t = 0 \) (initial time), the other is \( 9.192 - 4.9t = 0 \). Solving for \( t \): \( 4.9t = 9.192 \), so \( t = \frac{9.192}{4.9} \approx 1.876 \, \text{s} \).

Answer:

\( \approx 1.88 \) (or more precisely \( \frac{13\sqrt{2}}{9.8} \approx 1.88 \))