Problem 2 - Master Resource Book in JEE Main Physics

Question

A tennis ball rolls off the top of a sister case way with a horizontal velocity \(u \mathrm{~ms}^{-1}\). If the steps are \(b\) metre wide and \(h\) meter high, the ball will hit the edge of the \(n\)th step, if (al \(n=\frac{2 h u}{g b^{2}}\) (b) \(n=\frac{2 h u^{2}}{g b^{2}}\) (c) \(n=\frac{2 h u^{2}}{g b}\) (d) \(n=\frac{h u^{2}}{g b^{2}}\)

Step-by-Step Solution

Answer

The correct answer is (b) \( n=\frac{2hu^2}{gb^2} \).

Step 1: Analyze the Scenario

The tennis ball is given an initial horizontal velocity \( u \) and rolls off a set of stairs. The steps have dimensions \( b \) and \( h \), and we need to calculate at which step, \( n \), the ball will hit the stairs.

Step 2: Calculate Time of Flight for Step

First, we need to determine how far the ball falls vertically for each step. If the ball is to hit the \( n\)th step edge, it will need to fall a total vertical distance of \( n imes h \).

Step 3: Equation for Vertical Motion

Using the equation for vertical motion under gravity, \( s = \frac{1}{2}gt^2 \), where \( s \) is vertical distance, \( g \) is acceleration due to gravity, and \( t \) is time, we have:\[ n \times h = \frac{1}{2}gt^2 \]

Step 4: Expression for Time \( t \)

From the vertical motion equation, solve for \( t \):\[ t = \sqrt{\frac{2nh}{g}} \]

Step 5: Horizontal Displacement Calculation

The horizontal displacement needed for the ball to reach the \( n\)th step is \( n \times b \). Using the horizontal motion equation, \( d = ut \), substitute \( t \) from Step 4.\[ n \times b = u \sqrt{\frac{2nh}{g}} \]

Step 6: Solving for \( n \)

Square both sides of the equation from Step 5 to eliminate the square root and solve for \( n \):\[ n^2 b^2 = u^2 \frac{2nh}{g} \]Re-arrange to find \( n \):\[ n = \frac{2hu^2}{gb^2} \]

Step 7: Choose the Correct Option

From the possible options, compare the expression \( n = \frac{2hu^2}{gb^2} \) with the given options. It matches option (b).

Key Concepts

Horizontal Velocity

In projectile motion, especially horizontal motion, the concept of horizontal velocity is crucial. When a tennis ball rolls off an edge with a horizontal velocity denoted by \( u \), it means that the ball possesses a constant speed in the horizontal direction.

The important part about horizontal velocity is that it does not change—once the ball leaves the edge, there is no acceleration acting on it horizontally (ignoring air resistance). This makes calculations simpler because the ball's horizontal velocity remains constant throughout the motion.

Some critical points to remember about horizontal velocity:

• It is responsible for moving the object forward.
• It remains constant unless acted upon by another force.
• The formula used to calculate horizontal displacement (\( d = ut \)) relies solely on this constant velocity.

This notion of constant horizontal motion allows us to predict where the ball will land as it rolls off the staircase.

Vertical Motion under Gravity

Vertical motion under gravity is what determines how long an object stays in the air. When a tennis ball drops off a ledge, gravity immediately starts pulling it downward, accelerating the object with a constant acceleration \( g \) which is approximately \( 9.81 \, \text{m/s}^2 \) on Earth.

In this problem, the ball falls vertically to cover a distance of \( n \times h \), where \( h \) is the height of each step and \( n \) is the step number it reaches. The equation we use here is the basic motion formula for vertical displacement under gravity: \[ s = \frac{1}{2}gt^2\]
Here, \( s \) stands for the vertical displacement or the total height fallen. This formula is derived from integrating acceleration due to gravity to find how far an object falls over time.

In practical terms, this ensures we understand that as long as an object is in free fall, gravity will uniformly increase its downfall velocity, helping us calculate time and impact point accurately.

Time of Flight

The time of flight refers to the total time the projectile is in motion before hitting a target or ground. For horizontal projectiles, this is influenced primarily by the height it drops from, due to gravity's influence.

For our tennis ball, to know how long it stays in the air until it hits the \( n \)th step, we calculate the time by rearranging the vertical motion equation. Solving \( n \times h = \frac{1}{2}gt^2 \) for time, we get:\[ t = \sqrt{\frac{2nh}{g}} \]

This equation clearly shows that the time of flight depends on both the number of steps \( n \) and the height of each step \( h \). The independence from horizontal velocity here is crucial because it highlights how vertical and horizontal motions are separate in projectile motion analysis.

Understanding the time of flight aids in predicting where and when the projectile will make contact, which is essential in strategic games involving projectiles.

Displacement Calculation

In projectile motion, displacement calculation plays a key role in determining exactly where the object lands. For horizontal motion, displacement is influenced only by the horizontal velocity and time of flight.

Using the formula \( d = ut \), the horizontal displacement we want is \( n \times b \) because it directly relates to the cumulative width of the steps the ball crosses. By setting the calculated time of flight into the horizontal motion formula, we find:\[ n \times b = u \cdot \sqrt{\frac{2nh}{g}} \]
Squaring both sides to remove the square root helps to solve for \( n \) and ensures a balanced equation. It is crucial because it precisely tells us how far the ball has traveled horizontally when it hits the \( n \)th step.

Understanding displacement rearranges our focus from just how far or fast, to exactly at which specific point the projectile impacts, thus pinning down the precise step it lands on in the staircase scenario.