A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3m/s. At what height, did he bail out?

After bailing out the parachutist travels 50m 
 

So  $v^2-u^2=2as$ $\dots \dots$(1)
using 1

$$\begin{gathered} v^2=2*9.8*50 \\ {v}^2=980. \end{gathered}$$

After the parachute is opened he decelerates at $2m{s}^{-2}$

by using 1

${\mathrm{v}}^2-3^2=2\left(2\right)*\mathrm{S}, \mathrm{S} \mathrm{is} \mathrm{the} \mathrm{distance} \mathrm{before} \mathrm{reaching} \mathrm{the} \mathrm{ground}$

S= 242.75m 

So total height =242.75+50 $\simeq$293m