Problem 36 - Master Resource Book in JEE Main Physics

Question

A ball is dropped from a bridge at a height of \(176.4 \mathrm{~m}\) over a river. After \(2 \mathrm{~s}\), a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously? (a) \(2.45 \mathrm{~ms}^{-1}\) (b) \(49 \mathrm{~ms}^{-1}\) (c) \(14.5 \mathrm{~ms}^{-1}\) (d) \(24.5 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The initial velocity of the second ball should be \(24.5 \mathrm{~ms}^{-1}\).

Step 1: Calculate Time for First Ball to Hit Water

Use the equation for the displacement of an object under gravity, which is given by \[ h = \frac{1}{2}gt^2 \] where \( h = 176.4 \mathrm{~m} \) is the height, \( g = 9.8 \mathrm{~ms}^{-2} \) is the acceleration due to gravity, and \( t \) is the time it takes. Solving for \( t \), we have:\[ 176.4 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 176.4 = 4.9t^2 \] \[ t^2 = \frac{176.4}{4.9} = 36 \]\[ t = 6 \mathrm{~s} \].The first ball takes 6 seconds to hit the water.

Step 2: Calculate Time Available for Second Ball

The second ball is thrown after 2 seconds, so it has less time to reach the water. Calculate this time by subtracting 2 seconds from the total time it takes for the first ball to hit the water:\[ t_\text{available} = 6 \mathrm{~s} - 2 \mathrm{~s} = 4 \mathrm{~s} \]. The second ball has 4 seconds to reach the water.

Step 3: Calculate Required Initial Velocity for Second Ball

Use the kinematic equation for the motion of the second ball: \[ h = v_0 t + \frac{1}{2}gt^2 \]Where \( h = 176.4 \mathrm{~m} \), \( t = 4 \mathrm{~s} \), and \( g = 9.8 \mathrm{~ms}^{-2} \). Solve for initial velocity \( v_0 \):\[ 176.4 = v_0 \times 4 + \frac{1}{2} \times 9.8 \times 4^2 \]\[ 176.4 = 4v_0 + 0.5 \times 9.8 \times 16 \] \[ 176.4 = 4v_0 + 78.4 \] \[ 4v_0 = 176.4 - 78.4 \]\[ 4v_0 = 98 \] \[ v_0 = \frac{98}{4} \] \[ v_0 = 24.5 \mathrm{~ms}^{-1} \] The initial velocity needed for the second ball is \( 24.5 \mathrm{~ms}^{-1} \).

Key Concepts

Kinematic Equations

Kinematic equations are fundamental tools in physics to describe the motion of objects. These equations relate the five main parameters of motion: displacement, initial velocity, final velocity, acceleration, and time. They are particularly useful for solving problems involving constant acceleration, such as projectile motion. In the context of the exercise, the equation \( h = v_0 t + \frac{1}{2}gt^2 \) is used to determine the initial velocity of the second ball. This equation incorporates:

• \( h \): the height or vertical distance the ball travels.
• \( v_0 \): the initial velocity, which is what we're solving for.
• \( g \): the acceleration due to gravity (9.8 \( \mathrm{ms}^{-2} \)).
• \( t \): the time period over which the second ball is in motion.

By understanding and applying kinematic equations, we can find unknown variables, predict motion, and understand how different factors affect the trajectory of an object.

Acceleration due to Gravity

Acceleration due to gravity is a critical concept in understanding projectile motion. It is the constant acceleration experienced by an object in free fall toward Earth and is denoted by \( g \). On Earth's surface, \( g \) is approximately \( 9.8 \ \mathrm{ms}^{-2} \). This constant acceleration impacts how objects move in a vertical direction.
The exercise demonstrates using \( g \) in the equation \( h = \frac{1}{2}gt^2 \) to calculate how long it takes for an object to reach the ground free falling from a specific height. Due to \( g \), objects accelerate as they fall, increasing their speed by \( 9.8 \ \mathrm{ms}^{-1} \) every second.
Understanding gravity's role allows us to predict how long it will take an object to fall a certain distance and helps in solving problems related to projectile motion, such as finding the time or initial velocity required for one object to meet another at a specific point.

Initial Velocity Calculation

Initial velocity calculation is a pivotal aspect of solving projectile motion problems when an object is launched or thrown. To find the initial velocity \( v_0 \) of the second ball in the exercise, we use the rearranged kinematic equation:

• Recognize the displacement \( h \) as the distance the ball travels.
• Identify other known values: time \( t = 4 \) seconds and \( g = 9.8 \ \mathrm{ms}^{-2} \).

Given that the total displacement and the time of descent are known, you substitute these values into the equation:\[176.4 = v_0 \times 4 + \frac{1}{2} \times 9.8 \times 16\]This equation simplifies to find \( v_0 \), the initial velocity required for the second ball to hit the water simultaneously with the first ball. Ensure calculations are precise, and confirm that each part of the equation is accurately represented.