Problem 61 - Master Resource Book in JEE Main Physics

Question

A stone thrown vertically upwards attains a maximum height of \(45 \mathrm{~m} .\) In what time the velocity of stone become equal to one-half the velocity of throw? (Given \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(2 \mathrm{~s}\) (b) \(1.5 \mathrm{~s}\) (c) \(1 \mathrm{~s}\) (d) \(0.5 \mathrm{~s}\)

Step-by-Step Solution

Answer

The time is 1.5 s, which corresponds to option (b).

Step 1: Identify Known Values

We are given that the maximum height reached by the stone is 45 m and the acceleration due to gravity is \( g = 10 \text{ m/s}^2 \). Let's denote the initial velocity of the stone as \( u \).

Step 2: Use Kinematic Equation to Find Initial Velocity

To find the initial velocity \( u \), we can use the kinematic equation for the maximum height:\[ v^2 = u^2 - 2gh \]where \( v = 0 \) at the maximum height and \( h = 45 \text{ m} \).Substituting the values, we have:\[ 0 = u^2 - 2 \times 10 \times 45 \]\[ u^2 = 900 \]\[ u = 30 \text{ m/s} \].

Step 3: Find Velocity as Half of Initial Velocity

We need the velocity \( v \) to be half the initial velocity \( u \). So,\[ v = \frac{u}{2} = \frac{30}{2} = 15 \text{ m/s} \].

Step 4: Use Kinematic Equation to Find Time

Now, use the velocity-time relation:\[ v = u - gt \]Substitute \( v = 15 \text{ m/s} \), \( u = 30 \text{ m/s} \), and \( g = 10 \text{ m/s}^2 \):\[ 15 = 30 - 10t \]\[ 10t = 15 \]\[ t = 1.5 \text{ s} \].

Step 5: Conclusion

The time taken for the stone's velocity to become half of its initial velocity is \( t = 1.5 \text{ s} \). Therefore, the correct choice is (b) \( 1.5 \text{ s} \).

Key Concepts

Projectile Motion

Projectile motion happens when an object is thrown or projected into the air and moves under the influence of gravity. This involves a free-falling object following a path called a trajectory. The motion has two components:

• Horizontal motion with constant velocity (assuming no air resistance).
• Vertical motion with uniform acceleration due to gravity.

Understanding these components helps predict where and when an object like a thrown stone will reach its maximum height and how long the entire motion will last.
For a stone thrown vertically, the motion simplifies. Vertically, it goes up until gravity stops its upward motion, and then it falls back down. At the maximum height, its vertical velocity is zero because it momentarily stops before falling back down.

Acceleration due to Gravity

Acceleration due to gravity is a key concept in projectile motion and in the motion of any freely falling object on Earth. It's the acceleration that Earth's gravity imparts on any object, usually denoted by the symbol \( g \), and its standard value is approximately \( 9.81 \text{ m/s}^2 \). However, in many problems, it's often rounded to \( 10 \text{ m/s}^2 \) for simplicity in calculations.
Gravity always acts downwards towards the center of the Earth and influences the entire path of the projectile from the moment it is released. When an object is projected upwards, gravity slows it down until it reaches its peak, then speeds it up as it falls back. This consistent force is what causes the symmetric motion of the ascent and descent in ideal projectile motion absence of air resistance.

Initial Velocity Calculation

Initial velocity is the speed and direction of an object at the moment it begins its motion. In the case of vertically thrown objects, calculating the initial velocity can be crucial as it determines how far and how high the object will go.
To find initial velocity, we can use the kinematic equation when an object reaches its maximum height. Since the maximum height reached is when the final velocity \( v \) is zero, the equation simplifies. For upward motion, we use:\[ v^2 = u^2 - 2gh\]where \( v = 0 \), \( g \) is the acceleration due to gravity, and \( h \) is the height reached. Substituting known values allows solving for \( u \), the initial velocity.
Understanding how to derive this initial speed helps predict future points of motion, like halfway velocities or the time it takes to get to certain speeds.