A football is kicked vertically upwards from the ground and a student gazing out of the window sees it moving upwards past her at \[5.00m/s\]. The window is \[15.0\]m above the ground. Air resistance may be ignored. Take \[g=10m/{{s}^{2}}\].
(a) How high does the football go above the ground?
(b) How much time does it take to go from ground to highest point?

Hint: Before you should solve the question, you should draw the diagram from the given information. So that it will be easier for us to solve the problem. The three equations of motion can be used to calculate the distance, velocity, position, and acceleration of the given object. At heights, the final velocity of the object becomes zero.

Complete answer:
It is shown in the below diagram that the window is located \[15.00m\] above the ground
So, \[s=15.00m\]

The three equations of motion are-
\[v=u+at\]……….(1)
\[{{v}^{2}}={{u}^{2}}+2as\]…….(2)
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]……….(3)
Where,
\[u=initial\text{ }velocity\]
\[v=final\text{ }velocity\]
\[t=time\]
\[s=dis\tan ce\]
\[a=acceleration\]

In the above question, it is given that a ball is thrown vertically upwards with a velocity \[5.00m/{{s}^{2}}\].
So the final velocity of the ball when going upwards will be \[v=5.00m/{{s}^{2}}\].
(a) In this part we have to find that how much distance does the ball covered in going upwards so for that total height of the ball should find out, which is as follows
 we know that according to the second equation of motion
\[{{v}^{2}}={{u}^{2}}+2as\] …………..Eq(4)
And acceleration will be \[a=-10m/{{s}^{2}}\]
It is given that
(Acceleration will be negative because the ball is moving against gravity)
It is also given that, \[s=15.0m\]
Putting all these values in Eq(4 ), we get
\[\begin{align}
  & {{v}^{2}}={{u}^{2}}+2as \\
 & \Rightarrow {{(5)}^{2}}={{u}^{2}}-2\times 10\times 15 \\
 & \Rightarrow 25={{u}^{2}}-300 \\
 & \Rightarrow 25+300={{u}^{2}} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{u}^{2}}=325 \\
 & \Rightarrow u=\sqrt{325}m/s \\
\end{align}\]
We know that the maximum height is given by –
\[H=\dfrac{{{u}^{2}}}{2g}\]………..Eq(5)

Where,
\[\begin{align}
  & H=maximum\text{ }height \\
 & u=initial\text{ }velocity \\
 & g=acceleration \\
\end{align}\]
So after putting the value of \[u=\sqrt{325}m/s\] in eq(5), we can find the maximum height, which is as follows
\[\begin{align}
  & H=\dfrac{{{u}^{2}}}{2g} \\
 & \Rightarrow H=\dfrac{325}{2\times 10} \\
\end{align}\]
\[\Rightarrow H=16.5m\]
So the maximum height of the ball thrown upwards will be \[H=16.5m\].
(b) In this part we have to find the time taken by the ball to go from the ground to its highest point
When the ball reaches the highest point then its final velocity will be zero.
So, from the first equation of motion ,
\[v=u+at\]
Where
\[\begin{align}
  & v=0m/s \\
 & u=\sqrt{325}m/s \\
 & a=-10m/{{s}^{2}} \\
\end{align}\]
On putting all these values in the above equation, we get
\[\begin{align}
  & \Rightarrow 0=\sqrt{325}-10\times t \\
 & \Rightarrow 10t=\sqrt{325} \\
 & \Rightarrow t=\dfrac{\sqrt{325}}{10} \\
 & \Rightarrow t=\dfrac{5\sqrt{13}}{10} \\
\end{align}\]
\[\Rightarrow t=\dfrac{\sqrt{13}}{2}\sec \]
\[\Rightarrow t=\dfrac{3.6}{2}\sec \]
(since the value of \[\sqrt{13}=3.6\])
\[\Rightarrow t=1.8\operatorname{s}\]
So the time taken by the ball to reach from ground to the highest point is \[t=1.8\operatorname{s}\].

Note: Retardation is another name for negative acceleration. Retardation is the opposite of acceleration which happens when the object moves opposite to the direction of velocity. Scalar quantities are the quantities that have only magnitude and no direction and vector quantities are the quantities that have both magnitude and direction.