+171 A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second.
1.Write an equation that fives the height (in feet) of the football as a function of the time (in seconds) sunce it was punted.
2. Find the height (in feet) of the football 2seconds after the punt
3 Caculate how many seconds after the ball would hit the ground.
Thanks
+118608 A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second.
1.Write an equation that fives the height (in feet) of the football as a function of the time (in seconds) sunce it was punted.
This depends on the angle that the ball is kicked at.
Let up be positive vertical y direction.
If you let the acute angle between the horizonal and the initial direction of flight then
The initial vertical velocity is 47sin(theta)
The initial horizonal velocity is 47cos(theta)
\(\ddot y=-32 feet/sec\quad that \;is\;from\;gravity\\ \dot y=-32t+47sin\theta\\ y=-16t^2+(47sin\theta)t+3\\~\\ \)
2. Find the height (in feet) of the football 2seconds after the punt
\(\text{When t=2}\\ y=-16*2^2+(47sin\theta)*2+3\\ y=-64+94sin\theta+3\\ y=-61+94sin\theta\;\; feet, \qquad \text{or 0 feet if this is negative}\)
3 Caculate how many seconds after the ball would hit the ground.
Find t when y=0
\(0=-16t^2+(47sin\theta)t+3\\ t=\frac{-47sin\theta\pm\sqrt{47^2sin^2\theta-4*-16*3}}{2*-16}\\ t=\frac{-47sin\theta\pm\sqrt{2209sin^2\theta+192}}{-32}\\ t=\frac{47sin\theta\pm\sqrt{2209sin^2\theta+192}}{32}\\ \theta \;\text{ is acute so}\\ t=\frac{47sin\theta+\sqrt{2209sin^2\theta+192}}{32}\;\;seconds\)
If the ball is kicked horizonally theta=0 and it will take 0.43 seconds to hit the ground.
If the ball is kicked vertically theta=90 degrees it will take 3.05 seconds to hit the ground.
