Question

A particle is thrown up with a certain velocity and at an angle $\theta$ with the horizontal . The variation of kinetic energy $\left(E\right)$ with time $\left(t\right)$ is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is C


Let the mass and initial velocity of the particle is $m$ and $v$ respectively.
Initial velocity in $x-$ direction = $v_x=v\cos \theta$
Initial velocity in $y-$ direction = $v_y=v\sin \theta$
The velocity of the particle at any time $t$ is given by
$v_x=v\cos \theta$
$v_y=v\sin \theta -gt\dots \left(i\right)$

The kinetic energy $\left(E\right)$ of the particle at any time $t$ is
$E=\frac{1}{2}m[\sqrt{v_x^2+v_y^2}{]}^2$
$\Rightarrow E=\frac{1}{2}m\left(v_x^2+v_y^2\right)$
$\Rightarrow E=\frac{1}{2}m\left[\right(v\cos \theta {)}^2+(v\sin \theta -gt{)}^2]$
$\Rightarrow E=\frac{1}{2}m\left[\right(v^2{\cos }^2\theta +v^2{\sin }^2\theta )+g^2{t}^2)-2vgt\sin \theta ]$
$\Rightarrow E=\frac{1}{2}m(v^2+g^2{t}^2-2vgt\sin \theta )$
$\Rightarrow E=\frac{1}{2}m{g}^2{t}^2-mvgt\sin \theta +\frac{1}{2}m{v}^2$

This equation is similar to the equation of quadratic equation $y=a{x}^2+bx+c$ which represent a parabola. Plotting kinetic energy on $y-$ axis and time on $x-$ axis.
Also, equation $a=\frac{1}{2}m{g}^2>0$, So, the parabola is open upward.

Hence, option(c) is correct.

Why this question? Concept: For any quadratic equation, $y=a{x}^2+bx+c,$ if $a>0$, parabola is open upward and if $a<0$, parabola is open downward.