Question

A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of $19.6m$, he opens the parachute and decelerates at the rate of $1m{s}^{-2}$. If he reaches the ground with a speed of $4.6m{s}^{-1}$, how long was he in air?

• A. $10s$
• B. $12s$
• C. $15s$
• D. $17s$

Answer 1

Answer: (D)

$17s$

Let $t_1$ be the time before opening of parachute.
Using $h=ut+\frac{1}{2}g{t}^2$, we get, $19.6=0+\frac{1}{2}9.8t_1^2\Rightarrow t_1=2s$
Taking $v_1$ as velocity attained after falling through $19.6m$
Using $v^2=u^2+2gh$, we get $v_1^2=0+2\left(9.8\right)\left(19.6\right)\Rightarrow v_1=19.6m/s$
Again taking $t_2$ as time taken after opening of parachute
Using $v=u+at,$ we get $4.6=19.6-\left(1\right)t_2\Rightarrow t_2=15s$
Total time $=t_1+t_2=2+15=17s$