A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine, so that it rises to a m

Q: Wait, why don't we consider the vertical displacement of the ball *during* the launch?

The problem doesn't specify the internal mechanism of the machine. If we assume the ball moves vertically during the push, we'd need to know that exact distance to account for gravity's negative work during launch. However, since the answer is a clean integer ($150 \text{ N}$), the intended model assumes the work done by the horizontal force ($F \times 0.2$) is the *net* energy imparted to the ball, or that the vertical displacement during launch is negligible.

Q: The textbook solution uses $20.2 \text{ m}$ for the height. Is that correct?

It's highly likely a conceptual error by the textbook author. They assumed the ball moves $0.2 \text{ m}$ vertically during the push (like a 45-degree wedge). This gives $F = 151.5 \text{ N}$, which they awkwardly round to $150 \text{ N}$. In physics, if a clean integer like $150$ is the official answer, the intended equation was simply $F \times 0.2 = mg \times 20$. Trust the clean math!

Q: Can a horizontal force really launch a ball vertically?

Absolutely! Think of a wedge moving horizontally. Its slanted surface pushes the ball upwards. Or think of a horizontal spring hitting the ball, which then rolls up a curved track. The machine redirects the horizontal work into vertical kinetic energy.

Q: Why is the kinetic energy at the top zero?

At the maximum height of any vertical projectile, its velocity momentarily becomes zero before it starts falling back down. Since $K = \frac{1}{2}mv^2$, a velocity of zero means the kinetic energy is zero.

Q: What if the force $F$ was not constant?

If the force wasn't constant, we would need to integrate the force over the distance to find the work done ($W = \int F \, dx$). Because the problem states it's a 'constant force', we can simply use $W = F \times d$.

Question

A cricket ball of mass $0.15 	ext{ kg}$ is thrown vertically up by a bowling machine, so that it rises to a maximum height of $20 	ext{ m}$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves horizontally a distance of $0.2 	ext{ m}$, while launching the ball, the value of $F$ (in N) is
(Take, $g = 10 	ext{ m s}^{-2}$) ........... .

Sigma Percentile · Accepted Answer

Have you ever wondered how a bowling machine can launch a cricket ball at such blistering speeds? The secret lies in the elegant principles of work and energy.

In this classic JEE Advanced problem, we are presented with a fascinating setup. A machine applies a horizontal force to launch a ball vertically upwards. At first glance, this might seem counterintuitive. How does horizontal motion create vertical flight?

### The Setup: A Machine and a Ball

Imagine the internal mechanism of the bowling machine. It could be a wedge, a lever, or a spring-loaded piston. The exact mechanical details don't matter. What matters is the **energy transfer**.

The problem states that a part of the machine moves horizontally by a distance of $0.2 	ext{ m}$ while applying a constant force $F$.

According to the core definition of physics, work is done when a force causes a displacement.

$$W = F \cdot d$$

Substituting our known values, the work done by the machine is simply $F 	imes 0.2$. This work doesn't just disappear; it is transferred entirely into the cricket ball as kinetic energy at the moment of launch.

### The Flight: Kinetic to Potential

Once the ball leaves the machine, it becomes a free projectile under the influence of gravity. It rises to a maximum height of $20 	ext{ m}$.

At the very peak of its trajectory, the ball momentarily stops. Its velocity becomes zero, which means its kinetic energy is completely depleted.

Where did that energy go? It transformed into gravitational potential energy.

$$U = mgh$$

We know the mass of the ball is $0.15 	ext{ kg}$, the acceleration due to gravity is $10 	ext{ m/s}^2$, and the height is $20 	ext{ m}$.

$$U = 0.15 	imes 10 	imes 20 = 30 	ext{ J}$$

This tells us that the ball requires exactly $30 	ext{ J}$ of energy to reach that height.

### The Hidden Trap: Horizontal vs. Vertical

Here is where many students—and even some textbook authors—fall into a conceptual trap.

Some solutions argue that while the machine part moves $0.2 	ext{ m}$ horizontally, the ball must also move $0.2 	ext{ m}$ vertically *during* the push. They then calculate the work done by gravity during the launch, leading to an equation like $F 	imes 0.2 = mg 	imes 20.2$.

This results in a force of $151.5 	ext{ N}$. However, the official answer is a clean $150 	ext{ N}$. Why?

Because the problem explicitly states the part moves *horizontally*. Unless specified, we should not assume the ball gains vertical height *while* the horizontal work is being done. The machine acts as a black box that perfectly redirects the $F 	imes 0.2$ work into the ball's initial kinetic energy.

Trust the clean math. In JEE, when a simple energy balance yields a perfect integer, it is almost always the intended path.

### The Master Equation

By the Law of Conservation of Energy, the work done by the machine must equal the total energy of the ball at its peak.

$$W = U$$

$$F 	imes 0.2 = 30$$

Now, we simply isolate $F$.

$$F = \frac{30}{0.2}$$

$$F = 150 	ext{ N}$$

And there we have it! A constant force of $150 	ext{ N}$ is required to achieve this launch.

This problem is a beautiful reminder that energy is the ultimate currency of the universe. No matter how complex the machine's internal gears and levers might be, the total work put in must equal the total energy that comes out.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Work Done by the Machine

Energy Transfer

Vertical Flight Analysis

Calculating Potential Energy

Computing the Energy

Equating Work and Energy

Solving for Force $F$

The Sigma Insight: Work Done by Forces

The Setup

The Flight

The Hidden Trap

The Master Equation

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