A ball of mass 2 kg 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 buring its motion between 0 s and 8 s, `hat(i)` is along the horizontal and `hat(j)` is along the vertical direction. (Take `g=10 m//s^(2)`)

Options

A. `-2.56` kJ
B. `0.52` kJ
C. 1.76 kJ
D. `-2.44` kJ

Answer

A

Explanation

Given, velocity of the ball, `v=(20 hat(i)+24 hat(j))` m/s and mass of the ball, `m=2` kg
From the equation of motion, when a ball is throuwn,
`v=u-g t` ...(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, charge in potential energy of the ball = change in kinetic energy of the ball
`implies Delta PE =1/2 m u^(2)-1/2 m v^(2)`
`=1/2 m(u^(2)-v^(2))`
`=2/2(u.u-v.v)`
`=[(20 hat(i)+24 hat(j)).(20 hat(j)+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
Hence, change in potential energy of the ball after `t=85` s is `-2.56` kJ.