Question

A parachute of mass $m$ starts coming down with a constant acceleration $a$. Determine the ballast mass to be released for the parachute to have an upward acceleration of the same magnitude. Neglect air drag.

• A. $\frac{2ma}{a+g}$
• B. $\frac{ma}{a-g}$
• C. $\frac{ma}{a+g}$
• D. $\frac{2ma}{g-a}$

Answer 1

The correct option is A $\frac{2ma}{a+g}$

If $B$ is the upward thrust on parachute then$,$

$ma=mg-B$-------------------$\left(1\right)$

If $m^{\prime }$ is the mass released from the system so that the parachute moves up with acceleration ${}^{\prime }{a}^{\prime }$ then$,$

$(m-m^{\prime })a=B-(m-m^{\prime })g$------------------$\left(2\right)$

adding $\left(1\right)$ and $\left(2\right)$

We get$,$ $2ma-m^{\prime }a=m^{\prime }g\Rightarrow m^{\prime }=\frac{2ma}{a+g}$

Hence,

option $\left(A\right)$ is correct answer.