A parachutist after bailing  out fall for 5 sec and when it open he descends with an deceleration of $1m/s^2$ and reach the ground with a speed of $3m/s$ .Find the height at which he was dropped.

till up to 5 sec  he is like a freely falling body
$$\begin{gathered} we know, S=Ut+\frac{1}{2}a{t}^2,\text{\hspace{0.17em}here displacement=S;initial velocity=U; time=t; acceleration=a;} \\ up to 5 sec it is like a freely falling body \\ S=0+\frac{1}{2}\left(g\right)5^2 \end{gathered}$$ 
$S=\frac{1}{2}\times 10\times 5\times 5=125m$
 $$\begin{gathered} The velocity parachutist gain at t=5sec \\ V=U+at \\ \Rightarrow V=0+10\times 5 \\ \Rightarrow V=50m/s \end{gathered}$$

$$\begin{gathered} After the parachute is open initial velocity U=50m/s;final velocityV=3m/s; \\ acceleration a=-1m/s^2\text{\hspace{0.17em}; displacement=\hspace{0.17em}\hspace{0.17em}}S=? \end{gathered}$$ 
 $$\begin{gathered} we know,V^2-U^2=2as \\ substitute the values in above equation \end{gathered}$$
$$\begin{gathered} 3^2-{50}^2=2\left(-1\right)S \\ \Rightarrow S=1245.5m \end{gathered}$$ 
Total height is $1245.5+125=1370.5m$