Discussion Forum

1)

A  ball of mass 2kg is thrown from a tall building with velocity, v=$(20 m/s)\hat{i}+(24 m/s)\hat{j}$ at time t=0 s. Change in the potential energy of the ball

after, t=8 s is ( The ball is assumed to be in air during its motion between 0s and 8 s , $\hat{i}$  is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g=10 m/s^{2}$)

Answer:

Option A

Explanation:

 Given , velocity of the ball , v=$(20 \hat{i}+24\hat{j})$ m/s and mass of the ball, m=2kg

 From the equation of motion , when a ball is thrown, 

  v=u-gt   .........(i)

 Substituting the given values in Eq.(i) , we get

 v= $(20 \hat{i} +24 \hat{j})-(10 \hat{j})8$

  = $20 \hat{i}+24 \hat{j}-80 \hat {j}= 20 \hat{i}-56 \hat{j}$ .....(ii)

 From the law of conservation of energy , change in potential energy of the ball= change in kinetic energy of the ball

 $\Rightarrow$    $\triangle PE$=   $\frac{1}{2}mu^{2}-\frac{1}{2}mv^{2}$

= $\frac{1}{2}m(u^{2}-v^{2})$

 $=\frac{2}{2}(u.u-v.v)$

  =$[(20 \hat{i}+24 \hat{j}).(20 \hat{i}+24 \hat{j})- [(20 \hat{i}-56 \hat{j}).(20 \hat{i}-56 \hat{j})]$

 = $(20)^{2}+(24)^{2}-(20)^{2}-(56)^{2}$

 = $(24)^{2}-(56)^{2}$= 576-3136

  =-2560=-2.56 kJ