> Q. An athlete throws the shot-put of mass $ 4text{ }kg $ with initial speed of $ 2.2text{ }m{{s}^{-1}} $ at $ {{41}^{o}} $ from a height of $ 1.3text{ }m $ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $ g=9.8m/{{s}^{-2}} $ ) – LIVE ANSWER TODAY

Q. An athlete throws the shot-put of mass $ 4text{ }kg $ with initial speed of $ 2.2text{ }m{{s}^{-1}} $ at $ {{41}^{o}} $ from a height of $ 1.3text{ }m $ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $ g=9.8m/{{s}^{-2}} $ )

In This post you will get answer of Q.
An athlete throws the shot-put of mass $ 4text{ }kg $ with initial speed of $ 2.2text{ }m{{s}^{-1}} $ at $ {{41}^{o}} $ from a height of $ 1.3text{ }m $ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $ g=9.8m/{{s}^{-2}} $ )

Solution:

As there is no air resistance and gravitatitfhal force is a conservative force, we can apply mechanical energy conservation.

$ Rightarrow $ $ {{(KE)}_{f}}={{(KE)}_{i}}+Change,,in,,PE $

$ =frac{1}{2}m{{V}^{2}}+mgh $ $ =frac{1}{2}times 4times {{(2.2)}^{2}}+4times (9.8)times (1.3) $

$ =9.68+50.96 $

$ =60.64approx 62.84,,J $

Leave a Comment