Answer:

Explanation:
The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

The work done on the ball due to drag is:


![W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]](https://tex.z-dn.net/?f=W_%7Bdrag%7D%20%3D%20%280.599%5C%2Ckg%29%5Ccdot%20%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20%282.18%5C%2Cm-3.10%5C%2Cm%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%280.599%5C%2Ckg%29%5Ccdot%20%5B%287.05%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%284.19%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5D)

Hi there! Lets see!
- m is mass, and its units are kg
- k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²
Therefore:
![\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bm%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5B%5Cdfrac%7Bkg%7D%7Bs%5E2%7D%5D%7D%7D%20%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%5Ccdot%20s%5E2%7D%20%3D%20%5Csqrt%7B%5Bs%5D%5E2%7D%20%3D%20s)
The period is given in seconds so the formula is dimensionally correct.
Answer:
Football will be in air for 2.8139 sec
Explanation:
We have given maximum height h = 9.7 meters
Angle of projection 
We know that maximum height is given by 
So 

u = 22.396 m/sec
Time of flight is given by
