A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s². At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the airplane is ____

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s². At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the airplane is _____ m.

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

A. $92.5$

B. $62.5$

C. $20$

D. $82.5$

Correct Answer: $92.5$

Explanation

The paratrooper initially performs free fall for $2\,s$ starting from rest.

$$ v = gt = 10 \times 2 = 20\,m/s $$

Distance fallen during free fall is

$$ s_1 = \frac{1}{2}gt^2 = \frac{1}{2}\times10\times(2)^2 = 20\,m $$

After opening the parachute, the paratrooper starts deaccelerating with acceleration $a = -3\,m/s^2$.

Using the equation

$$ v^2 = u^2 + 2as $$

Here $u = 20\,m/s$, $v = 5\,m/s$.

$$ (5)^2 = (20)^2 - 2\times3\times s_2 $$

$$ 25 = 400 - 6s_2 $$

$$ s_2 = 62.5\,m $$

This distance brings the paratrooper to $10\,m$ above the ground.

Hence total initial height of the airplane is

$$ h = s_1 + s_2 + 10 = 20 + 62.5 + 10 = 92.5\,m $$

Explanation

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