Question

A balloon is at a height of 81 m is ascending upward with a velocity of 12 ms1. A body of 2 kg weight is dropped from it. If g = 10 ms2, the body will reach the surface of earth in
A) 1.5 sec B) 4.0 sec
C) 5.4 sec D) 6.7 sec

Answer 1

\[h=h_0+v_0t-\frac{ 1 }{ 2 }g·t^2\].Surface will be hit when h=0, then solve:\[h_0+v_0t-\frac{ 1 }{ 2 }g·t^2=0\rightarrow 81+12t-5t^2=0\]Solutions are t=-3 and t=5.4. Definitely, for obvious reasons we have to discard t=-3, then the solution is t=5.4

Answer 2

Another way is finding anergy balance:
The total initial energy is potential (mgh) plus kinetic (1/2mv^2)
When it hits the ground, it is only kinetic, then:\[mgh_0+\frac{ 1 }{ 2 }mv_0^2=\frac{ 1 }{ 2 }mv_f^2 \rightarrow v_f=\sqrt{2gh_0+v_0^2}=\sqrt{1764}=42\]Initial velocity is upwards (we will consider it is positive) and Vf (velocity when hitting the ground is negative) is downwards, then we need to find t in:
\[-v_f=v_0-g·t \rightarrow t=\frac{ v_0+v_f }{ g }=\frac{ 12+42 }{ 10 }=5.4\]

Answer 3

in the figure shown blocks A and B move with velocities v1 and v2 along horizontal direction.Find the ratio of v1 and v2