Respuesta :
Answer:
(a). H = 8.83 m
(b). Rising
Explanation:
Given,
- Initial velocity of the ball = u = 22.6 m/s
- Height of the crossbar = h = 3.05 m
- Distance of the crossbar from the initial position of the ball = r = 36.0 m
- Angle of projection = [tex]\theta\ =\ 47.0^o[/tex]
- Initial horizontal velocity of the ball =[tex]u_x\ =\ ucos\theta[/tex]
- Initial vertical velocity of the ball = [tex]u_y\ =\ usin\theta[/tex]
part (a)
In the horizontal direction,
Ball is moving with the constant initial velocity of [tex]u_x[/tex]
Let 't' be the time of the ball to reach at the crossbar of distance r.
[tex]\therefore r = u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{22.6\times cos47^o}\\\Rightarrow t\ =\ 2.33\ sec[/tex]
At time 't', let 'y' be the vertical displacement attained by the ball.
From the equation of kinematics,
[tex]y\ =\ u_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ 22.6sin47^0\times 2.33\ -\ 0.5\times 9.81\times 2.33^2\\\Rightarrow y\ =\ 11.88\ m[/tex]
Total distance of the ball above the crossbar = H = y - h = 11.88 - 3.05 = 8.83 m
part (b)
At the maximum height, vertical velocity of the ball becomes zero,
Let h be the maximum height attained by the ball.
From the kinematics,
[tex]\therefore (v_y)^2\ =\ (u_y)^2\ -\ 2gh\\\Rightarrow 2gh\ =\ (u_y)^2\\\Rightarrow h\ =\ \dfrac{(u_y)^2}{2g}\\\Rightarrow h\ =\ \dfrac{(22.6\times sin47^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 13.92\ m[/tex]
Maximum height attained by the ball with the given initial velocity is 13.92 m but at the crossbar it attains 11.88 m. Hence the ball will still rising approaching the crossbar.