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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.6 m/s at an angle of 45.0° to the horizontal. (a) By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)
_______m

(b) Does the ball approach the crossbar (and cross above or beneath it) while still rising or while falling?
rising or falling? _________

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.