Solution:
Solution
Let the speed with which the particle was thrown be $$u$$
So,
$$v_x=u \cos\theta$$
$$v_y= u\sin\theta-gt$$
$$x=u\cos\theta t$$
$$y=u\sin\theta t-\dfrac{1} {2} gt^2$$
Let $$v$$ be the resultant velocity at any time $$t$$
$$\Rightarrow v^2=v_x^2+v_y^2$$
$$\Rightarrow v^2=u^2-2ugt\sin\theta+g^2t^2$$
$$\Rightarrow v^2=u^2-2gy$$
Now, $$KE\alpha v^2$$ [$$\because$$ mass of particle is constant]
$$\Rightarrow KE\alpha u^2-2gy$$
This means the graph between $$KE$$ and $$y$$ will be a straight line with a negative slope.
Hence the graph shown in figure $$A$$ is incorrect.
Note- Equation for all other graphs can be easily obtained by substituting the equations developed above
Correct answer is option 1