A paratrooper jumps from an aeroplane and opens a parachute after $2\ \mathrm{s}$ of free fall and starts deaccelerating with $3\ \mathrm{m/s^2}$. At $10\ \mathrm{m}$ height from ground, while descending with the help of parachute, the speed of paratrooper is $5\ \mathrm{m/s}$. The initial height of the airplane is ____ m.

$$
\left(g=10\ \mathrm{m/s^2}\right)
$$

Question

A paratrooper jumps from an aeroplane and opens a parachute after $2\ \mathrm{s}$ of free fall and starts deaccelerating with $3\ \mathrm{m/s^2}$. At $10\ \mathrm{m}$ height from ground, while descending with the help of parachute, the speed of paratrooper is $5\ \mathrm{m/s}$. The initial height of the airplane is ____ m.

$$
\left(g=10\ \mathrm{m/s^2}\right)
$$

Accepted Answer

A. 1. Divide the motion into **two intervals**.

* **Interval 1:** Free fall for $2\ \mathrm{s}$.
   * **Interval 2:** Motion after the parachute opens with a constant deacceleration of $3\ \mathrm{m/s^2}$.

2. **Interval 1: Free fall**

The paratrooper starts from rest:
   $$
   u_1=0,\qquad t_1=2\ \mathrm{s}
   $$

Distance covered:
   $$
   h_1=u_1t_1+\frac{1}{2}gt_1^2
   $$
   $$
   =0+\frac{1}{2}	imes10	imes2^2
   =20\ \mathrm{m}
   $$

Velocity at the end of free fall:
   $$
   v_1=u_1+gt_1
   $$
   $$
   =0+10	imes2
   =20\ \mathrm{m/s}
   $$

3. **Interval 2: After opening the parachute**

Initial velocity:
   $$
   u_2=20\ \mathrm{m/s}
   $$

Final velocity at a height of $10\ \mathrm{m}$ above the ground:
   $$
   v_2=5\ \mathrm{m/s}
   $$

Acceleration:
   $$
   a=-3\ \mathrm{m/s^2}
   $$

Using the third equation of motion:
   $$
   v_2^2=u_2^2+2ah_2
   $$

Substituting the values:
   $$
   5^2=20^2+2(-3)h_2
   $$
   $$
   25=400-6h_2
   $$
   $$
   h_2=\frac{375}{6}=62.5\ \mathrm{m}
   $$

4. The total initial height of the airplane is:
   $$
   H=h_1+h_2+10
   $$
   $$
   =20+62.5+10
   =92.5\ \mathrm{m}
   $$

5. Therefore, the airplane was initially at a height of **$92.5\ \mathrm{m}$**, so the correct option is **A**.