Given:
The initial speed of teh ball =62 feet per second.
The angle is 15 degrees.
The height is 2.2 feet.
Required:
We need to find the distance from the start point to the end that is the ball land on the ground.
Explanation:
[tex]v_0=62,\theta=15\degree,h=2.2[/tex]We need to find the horizontal position.
Consider the horizontal position equation.
[tex]x=(v_0cos\theta)t[/tex][tex]Substitute\text{ }v_0=62,\text{ and}\theta=15\degree\text{ in the equation}[/tex][tex]x=(62cos15\degree)t[/tex]We need to find the time traveled by the ball in the air to find the distance.
Consider the vertical position equation.
[tex]y=h+(v_0sin\theta)t-16t^2[/tex][tex]Substitute\text{ }v_0=62,\text{ }h=2.2,\text{ and}\theta=15\degree\text{ in the equation}[/tex][tex]y=2.2+(62sin15\degree)t-16t^2[/tex]Set y =0 and solve for t.
[tex]0=2.2+(62sin15\degree)t-16t^2[/tex][tex]0=2.2+16.0468t-16t^2[/tex][tex]-16t^2+16.0468t+2.2=0[/tex]Use quadratic formula.
[tex]t=\frac{-16.0468\pm\sqrt{(16.0468)^2-4(-16)(2.2)}}{2(-16)}[/tex][tex]t=\frac{-16.0468\pm19.9574}{-32}[/tex][tex]t=\frac{-16.0468+19.9574}{-32},\frac{-16.0468-19.957,4}{-32}[/tex][tex]t=-0.1222,1.1251[/tex]The value of time is always positive.
We get t =1.1251.
[tex]x=(62cos15\degree)t[/tex]Substitute t =1.1251 in the vertical position equation.
[tex]x=(62cos15\degree)\times1.1251[/tex][tex]x=67.3793feet[/tex]Final answer:
The ball travels approximately 67.379 feet.
