Q2.7-48PE
Question
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of \(11.0 m/s\)How long does he have to get out of the way if the shot was released at a height of\(2.20 m\), and he is \({\bf{1}}.{\bf{80}}{\rm{ }}{\bf{m}}\) tall?
Step-by-Step Solution
VerifiedTime taken to get out of the way if the shot was released is \(2.27 S\)
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 taken to 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 taken to get out of the way if the shot was released is \(2.27 S\)