Respuesta :
Answer:
Part (a) 10.15 m
Part (b) Rising
Explanation:
Given,
- Initial speed of the ball = u = 23.6 m/s
- Height of the crossbar = h = 3.05 m
- Distance between the ball and the cross bar = r = 36.0 m
- Angle of projection = [tex]\theta\ =\ 45.0^o[/tex]
- Initial velocity of the ball in the horizontal direction = [tex]u_x\ =\ ucos\theta[/tex]
- Initial velocity of the ball in the vertical direction = [tex]u_y\ =\ usin\theta[/tex]
part (a)
Let 't' be the time taken to reach the ball to the cross bar,
In x-direction,
[tex]\therefore r\ =\ u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{23.6cos45^o}\\\Rightarrow t\ =\ 2.15\ sec[/tex]
Let y be the height attained by the ball at time t = 2.15 sec,
[tex]y\ =\ u_yt\ \ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ 23.6\times sin45^o\times 2.15\ -\ 0.5\times 9.81\ 2.15^2\\\Rightarrow y\ =\ 13.205\ m[/tex]
Now Let H be the height by which the ball is clear the crossbar.
[tex]\therefore H\ =\ y\ -\ h\ =\ 13.205\ -\ 3.05\ =\ 10.15\ m[/tex]
part (b)
At the maximum height the vertical velocity of the ball becomes zero.
i,e, [tex]v_y\ =\ 0[/tex]
Let h be the maximum height attained by the ball.
[tex]\therefore v_y^2\ =\ u_y^2\ -\ 2gh\\\Rightarrow 0\ =\ (usin\theta)^2\ -\ 2gh\\\Rightarrow h\ =\ \dfrac{(usin\theta)^2}{2g}\\\Rightarrow h\ =\ \dfrac{23.6\times sin45.0^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 14.19\ m[/tex]
Hence at the cross bar the ball attains the height 13.205 m but the maximum height is 14.19 m. Therefore the ball is rising when it reaches at the crossbar.