Problem 51 - Master Resource Book in JEE Main Physics

Question

A parachutist after alling out falls \(50 \mathrm{~m}\) without friction. When parachute opens, it decelerates at \(2 \mathrm{~ms}^{-2}\). He reaches the ground with a speed of \(3 \mathrm{~ms}^{-1}\). At what height, did he fallen out? [AIEEE 2005] (a) \(111 \mathrm{~m}\) (b) \(293 \mathrm{~m}\) (c) \(182 \mathrm{~m}\) (d) \(91 \mathrm{~m}\)

Step-by-Step Solution

Answer

The correct height is approximately 200m, not matching any options; check problem or answer options.

Step 1: Calculate Speed Before Parachute Opens

First, find the speed of the parachutist after free falling 50m without air resistance using the equation for motion: \( v^2 = u^2 + 2as \), where \(u = 0\), \(a = 9.81\, m/s^2\), and \(s = 50\, m\). This gives \( v^2 = 0 + 2 \times 9.81 \times 50 \). Thus, \( v = \sqrt{981} \approx 31.3 \, m/s \).

Step 2: Apply Deceleration with Open Parachute

Use the final speed after deceleration with the equation \( v^2 = u^2 + 2as \), where the final speed \(v = 3\, m/s\), initial speed \(u = 31.3\, m/s\), and acceleration \(a = -2\, m/s^2\). Plugging in these values gives \( 9 = (31.3)^2 - 4s \), solving for \(s\) gives 150.31m.

Step 3: Calculate Total Height

The total height fallen is the sum of the initial 50m free fall and the 150.31m during deceleration: \( \text{Total Height} = 50\, m + 150.31\, m = 200.31\, m \).

Key Concepts

Free Fall

Free fall is a type of motion where an object is moving only under the influence of gravity, meaning there is no air resistance acting on it. In these conditions, the object accelerates downwards at a constant rate, known as the gravitational acceleration. On Earth, this acceleration is approximately \[ 9.81 \, \text{m/s}^2 \] This value can vary slightly depending on where you are on the planet, but for most calculations, it is accepted as 9.81 m/s². In the parachutist's scenario, the initial free fall before the parachute opened was considered to have no air resistance. Therefore, his motion was determined solely by gravity.

• The initial speed \( u \) of the parachutist during free fall is 0, as he starts from a standstill.
• Using the speed equation \( v^2 = u^2 + 2as \) helps to calculate the speed after falling a certain distance.

The parachutist fell 50 meters under these conditions, reaching a speed calculated by substituting into the formula: \[ v = \sqrt{2 \times 9.81 \times 50} \approx 31.3 \, \text{m/s} \] This demonstrates how fast something can fall when unaffected by anything other than gravity itself.

Deceleration

Deceleration occurs when an object slows down, meaning its velocity decreases over time. In physical terms, this is referred to as negative acceleration. In the case of our parachutist, deceleration happened when the parachute opened.When the parachute opens, it significantly increases the air resistance. This newly introduced force opposes the gravitational pull, thereby reducing the speed at which the parachutist falls. In this problem, the deceleration was given as \[ -2 \, \text{m/s}^2 \]This value of deceleration is factored into the motion equation to determine how much further the parachutist travels before coming down to the ground at a safe speed.

• Initial speed before deceleration \( u \) is obtained from the free fall: 31.3 m/s.
• Final speed \( v \) when reaching the ground is 3 m/s.
• Deceleration \( a \) is -2 m/s², which indicates slowing down.

Using the motion equation \( v^2 = u^2 + 2as \) and solving for distance \( s \), we can find: \[ 9 = (31.3)^2 + 2 \times (-2) \times s \] This simplifies to give a further traveled distance of approximately 150.31 meters before reaching the ground safely.

Motion Equations

Motion equations, also known as the equations of kinematics, help describe the motion of objects. They connect displacement, velocity, acceleration, and time in a mathematical manner. There are three primary equations of motion used for calculations:1. \( v = u + at \) — relates velocity to time when acceleration is constant.2. \( s = ut + \frac{1}{2}at^2 \) — connects displacement to time.3. \( v^2 = u^2 + 2as \) — links velocity with displacement.In this parachutist problem, the third equation was critical. It allows the calculation of speed and displacement without needing the time dimension.

• First, we used it to find the speed after the free fall (50m without parachute): \( v^2 = u^2 + 2as \) Translated to: \( 31.3^2 = 0 + 2 \times 9.81 \times 50 \)
• Second, to determine the additional distance covered after deceleration with the parachute: \( 9 = (31.3)^2 + 2 \times (-2) \times s \).

Motion equations offer a systematic way to approach problems of moving bodies, making them incredibly useful in a wide range of physics applications.