Question

An aeroplane drops a parachutist. After covering a distance of $40m$, he opens the parachute and retards at $2m{s}^{-2}$. If he reaches the ground with a speed of $2m{s}^{-1}$, he remains in the air for about

• A. $16s$
• B. $3s$
• C. $13s$
• D. $10s$

Answer 1

The correct option is A $16s$

Let $t_1$ be the time before opening of parachute.
Using $h=ut+\frac{1}{2}g{t}^2$, we get, $40=0+\frac{1}{2}9.8t_1^2\Rightarrow t_1=3s$
Taking $v_1$ as velocity attained after falling through $40m$
Using $v^2=u^2+2gh$, we get $v_1^2=0+2\left(9.8\right)\left(40\right)\Rightarrow v_1=28m/s$
Again taking $t_2$ as time taken after opening of parachute
Using $v=u+at,$ we get $2=28-\left(2\right)t_2\Rightarrow t_2=13s$
Total time $=t_1+t_2=3+13=16s$