Answer:
The distance by the ball clear the crossbar is 1.15 m
Explanation:
Given that,
Distance = 44 m
Speed = 24 m/s
Angle = 31°
Height = 3.05 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity
![u_{x}=u\cos\theta](https://tex.z-dn.net/?f=u_%7Bx%7D%3Du%5Ccos%5Ctheta)
Put the value into the formula
![u_{x}=24\cos(31)](https://tex.z-dn.net/?f=u_%7Bx%7D%3D24%5Ccos%2831%29)
![u_{x}=20.5\ m/s](https://tex.z-dn.net/?f=u_%7Bx%7D%3D20.5%5C%20m%2Fs)
We need to calculate the vertical velocity
Using formula of vertical velocity
![u_{y}=u\sin\theta](https://tex.z-dn.net/?f=u_%7By%7D%3Du%5Csin%5Ctheta)
Put the value into the formula
![u_{y}=24\sin(31)](https://tex.z-dn.net/?f=u_%7By%7D%3D24%5Csin%2831%29)
![u_{y}=12.3\ m/s](https://tex.z-dn.net/?f=u_%7By%7D%3D12.3%5C%20m%2Fs)
We need to calculate the time
Using formula of time
![t=\dfrac{d}{u_{x}}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bu_%7Bx%7D%7D)
Put the value into the formula
![t=\dfrac{44}{20.5}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B44%7D%7B20.5%7D)
![t=2.1\ sec](https://tex.z-dn.net/?f=t%3D2.1%5C%20sec)
We need to calculate the vertical height
Using equation of motion
![h=u_{y}t+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Du_%7By%7Dt%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Put the value into the formula
![h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2](https://tex.z-dn.net/?f=h%3D12.3%5Ctimes2.1-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%282.1%29%5E2)
![h=4.2\ m](https://tex.z-dn.net/?f=h%3D4.2%5C%20m)
We need to calculate the distance by the ball clear the crossbar
Using formula for vertical distance
![d=h-3.05](https://tex.z-dn.net/?f=d%3Dh-3.05)
Put the value of h
![d=4.2-3.05](https://tex.z-dn.net/?f=d%3D4.2-3.05)
![d=1.15\ m](https://tex.z-dn.net/?f=d%3D1.15%5C%20m)
Hence, The distance by the ball clear the crossbar is 1.15 m