Question

A ball of mass $m=1Kg$ is thrown from the top of a building with initial velocity $\vec{v}=(20m/s)\hat{i}+(24m/s)\hat{j}$ at time $t=0s$. The change in the potential energy of the ball between $t=0$ and $t=6s$, if the ball does not hit the ground
(Assume $g=10m/s^2$ )

• A. $-320J$
• B. $-360J$
• C. $-380J$
• D. $+320J$

Answer 1

Answer: (B)

$u=20\hat{i}+24\hat{j}$

To find change in PE, we have to consider only vertical motion of the ball.
So, we have
$u_y=24m/s,m=1kg$
$a_y=-10m/s^2,t=6s$
We take origin at point of projection,
$h=u_{y}t-\frac{1}{2}g{t}^2=24\times 6-\frac{1}{2}\times 10\times 36$
$=144-180=-36m$
So, vertical displacement of the particle is
$h=-36m$
Hence, change in $PE=mgh$
$=1\times 10\times -36=-360J$