Answer:
1.42 m
Step-by-step explanation:
Let's first write down the data given
We have
![\Delta{$x$} $ = horizontal distance from the kickoff point to goal = 24 $m](https://img.qammunity.org/2023/formulas/physics/college/psdisc5ilhsnirq4grmie7wxmhwqd8iqrr.png)
![h$ = height of the crossbar = 3.05 m](https://img.qammunity.org/2023/formulas/physics/college/88jwgm0jsnhrhs3fcmefve7v8th0cssnip.png)
![\theta$ = angle at which the ball leaves the ground = 53,8^(\circ)](https://img.qammunity.org/2023/formulas/physics/college/1pxwrm9g3izawsl4nq75mx09fkjq0wrq6w.png)
![v = 17 m/s](https://img.qammunity.org/2023/formulas/physics/college/p018w2iw5x7tclk40aabzmql1ymftbev2p.png)
.
Note that we take g to -9.8 m/s² since gravity is acting in the opposite direction to the vertical movement of the ball. Thus the ball is slowing down with increasing height
The horizontal component of the velocity
![v_x= $v\cos(53.8) = 17 (0.59) = 10.04 $ m/s](https://img.qammunity.org/2023/formulas/physics/college/3lfrj2aoekc4u5rs6w9f2u241siob5c62m.png)
Time t, taken to traverse distance of 24m = Δx/vx = 24/10.04 = 2.39s
The vertical component of velocity
The vertical displacement is given by
![\Delta y = v_yt + (1)/(2)gt^2\\\\= 13.58 * 2.39 - (1)/(2) * 9.8 * 2.39^2\\\\= 4.46691 m](https://img.qammunity.org/2023/formulas/physics/college/5kxee3ynm2c4m02wy02t3y53tl6h6m0h2p.png)
Since the height of the cross bar is 3.05m, the ball clears the crossbar by 4.46691 - 3.05 = 1.41691 ≈ 1.42 m