Question

A parachutist after bailing out falls 80 m without air resistance. Then the parachute is opened, and he decelerates at a uniform rate of $2m/s^2$. If he reaches the ground with a speed of $4m/s$, then the height at which he bailed out is [$g=10m/s^2$]

• A. $576m$
• B. $676m$
• C. $476m$
• D. $376m$

Answer 1

The correct option is C $476m$

Final velocity after falling 80m,

$V_s^2=0+2\times 10\times 80(v_c=0)$

$V_s=\sqrt{1600}$
$V_{s1}=40m/s$

Final velocity when reaching ground =4 m/s

$(V_{s2}{)}^2=V_{s1}^2-2ah$

$4^2={40}^2-4h$

$h=\frac{(40-4)(40+4)}{4}$

$h=9\times 44=396$

Total height of bailing $=396+80=476m$