Problem 106 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A body falling from rest under gravity passes a certain point \(P\). It was at a distance of \(400 \mathrm{~m}\) from \(P, 4\) s prior to passing through \(P\). If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), then the height above the point \(P\) from where the body began to fall is [2006] (a) \(720 \mathrm{~m}\) (b) \(900 \mathrm{~m}\) (c) \(320 \mathrm{~m}\) (d) \(680 \mathrm{~m}\)

Step-by-Step Solution

Answer

The height from where the body began to fall is 480 m. Answer: (d) 680 m.

Step 1: Identify Known Values

We know that the initial velocity \( u = 0 \) (since the body falls from rest), the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), and the distance to the point \( P \), \( s = 400 \, \text{m} \), which was covered in \( t = 4 \, \text{s} \).

Step 2: Use Kinematic Equation for Distance

We use the kinematic equation for distance covered under constant acceleration: \[ s = ut + \frac{1}{2} g t^2 \]. Substitute the known values: \( u = 0 \), \( g = 10 \), and \( t = 4 \). This gives \[400 = 0 \times 4 + \frac{1}{2} \times 10 \times (4)^2\].

Step 3: Solve the Equation

Calculate the expression on the right side: \[400 = \frac{1}{2} \times 10 \times 16 = 80 \, \text{m}\]. This matches the distance just below the height from where the body actually started the fall.

Step 4: Determine Height Above P

Since the body fell an additional 80 m to reach the point \( P \), the total initial height above point \( P \) is the sum of the two distances: \[ h = 400 + 80 = 480 \, \text{m} \].

Key Concepts

Motion Under Gravity

When understanding motion under gravity, it's important to remember that any object dropped from rest freely falls under the influence of gravitational force. The force brings the object towards the Earth without any other influence, like air resistance, being significant during the motion. Gravity accelerates the object at a constant rate, denoted as "g," typically valued at approximately 9.81 m/s² for Earth. However, in some exercises, such as the one we are discussing, a simplified value of 10 m/s² is often used for easy calculations.

Motion under gravity is a classic example of uniformly accelerated motion, where the acceleration remains consistent throughout the fall. An important aspect is that the initial velocity, usually represented by "u," is zero if the object starts from rest. As the object continues to fall, its velocity increases due to the constant gravitational acceleration. By understanding these basic principles, students can tackle a variety of physics problems that involve freely falling objects.

Distance and Velocity

In the context of kinematics, distance and velocity are two foundational concepts. When an object falls under gravity, the distance it covers continuously increases over time, while its velocity also changes. Velocity is the speed of the object, but with direction, making it a vector quantity.

For an object falling from rest, distance

• Starts at zero when the object begins its fall, and
• Increases quadratically over time.

Similarly, velocity at any point in time during free fall can be calculated using the kinematic equation \[v = u + gt\]where "u" is the initial velocity (zero if starting from rest) and "t" is the time elapsed. Hence, velocity is linearly proportional to time in this scenario.

Understanding the relationship between distance and velocity, especially in gravity's context, allows you to predict how far and how fast an object will move over any given period. This knowledge is not only crucial for exams but also practical physics applications like engineering and safety design.

Kinematic Equations

Kinematic equations are essential tools in physics that describe the motion of objects. They allow us to predict future positions and velocities based on initial conditions and constant accelerations, like gravity.

The primary kinematic equation used for distance under constant acceleration is:
\[s = ut + \frac{1}{2} g t^2\]This equation tells us the distance "s" an object travels over time "t", given its initial velocity "u" and the constant acceleration "g". Since the initial velocity is zero for a body falling from rest, the equation simplifies to:
\[s = \frac{1}{2} g t^2\]This simplification makes calculations more straightforward by removing the first term of the equation.

Another frequent equation, used to determine velocity at any time during the fall, is:
\[v = u + gt\]This shows that velocity increases linearly with time, a direct result of constant acceleration. In problems involving free fall, employing these equations helps calculate unknowns like distance traveled or final velocity, facilitating a deeper understanding of motion mechanics.