91Ó°ÊÓ

The above problem can be solved with the help of free-falling body

Here the shot put is been thrown upwards and then it comes downwards so, we have to calculate the total amount of the time player has to move.

Let’s see the sketch to understand the situation more easily

Hence the above problem can be divided into two parts

Part A: when the shot put goes up

Part B: when the shot put goes down

Given data for the part A section is as below:

U=11 m/s

g= -9.81 m/s²

V= 0

\(\begin{array}{l}v = u + gt\\0 = 11 + \left( { - 9.81} \right)t\\{t_A} = 1.12\,\,s\end{array}\)

Hence the time taken by the shot put to go till the top is 1.12 s

Distance covered by the shotput

\(\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( 0 \right)^2} - {\left( {11} \right)^2} = 2\left( { - 9.81} \right)d\\d = 6.167\,\,m\end{array}\)

Hence the total distance covered during the upward motion is 6.167 m

Now when the shot put falls down on the head of the person the total distance covered is the sum of two distance

\(\begin{array}{l}D = 6.167 + 0.4\\D = 6.57\,m\end{array}\)

The time calculated by the shot put in the part B is given by using following data

U=0

g=9.81 m/s²

d=6.57 m

\(\begin{array}{c}d = ut + \frac{1}{2}g{t^2}\\6.57 = \left( 0 \right)\left( t \right) + \frac{1}{2}\left( {9.81} \right){\left( t \right)^2}\\t = \,1.157\,\,s\end{array}\)

Hence, the time taken in part B is \(1.157{\rm{ }}s\)

The time takento get out of the way if the shot was released is

\(\begin{array}{l}T = 1.157 + 1.12\\T = 2.277\,\,\,s\end{array}\)

Therefore, time takento get out of the way if the shot was released is\(2.27 S\)