Question

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height. What is the direction and magnitude of the ball's velocity 0.2 seconds after impact?

Answer 1

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$, so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$.

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$