Problem 55 - Master Resource Book in JEE Main Physics

Question

A ball is thrown vertically upwards from the top of a tower of height \(h\) with velocity \(v\). The ball strikes the ground after $$ \begin{array}{ll} \text { (a) } \frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^{2}}}\right] & \text { (b) } \frac{v}{g}\left[1+\sqrt{1-\frac{2 g h}{v^{2}}}\right] \\ \text { (c) } \frac{v}{g}\left(1+\frac{2 g h}{v^{2}}\right)^{1 / 2} & \text { (d) } \frac{v}{g}\left(1-\frac{2 g h}{v^{2}}\right)^{1 / 2} \end{array} $$

Step-by-Step Solution

Answer

Option (a) is correct: \( \frac{v}{g} \left[ 1 + \sqrt{1+\frac{2gh}{v^2}} \right] \).

Step 1: Define the Kinematic Equation

The kinematic equation for an object moving under gravity is given by \( s = ut + \frac{1}{2} a t^2 \), where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. For the ball thrown upwards, \( u = v \) (the initial velocity), \( a = -g \) (acceleration due to gravity), and \( s = -h \) because the ball ends below the tower's top.

Step 2: Substitute Values into the Equation

Substituting the known values into the kinematic equation: \(-h = vt - \frac{1}{2}gt^2\). This equation describes the motion of the ball from top of the tower until it strikes the ground.

Step 3: Define the Time Up and Down Movement

The ball first goes up, reaches its maximum height, and then falls to the ground. The time \( t_1 \) to reach maximum height is given by \( t_1 = \frac{v}{g} \). The total time \( T \) is then given by \( T = t_1 + t_2 \), where \( t_2 \) is the time to descend from the maximum height to the ground.

Step 4: Solve for Time of Descent

The descent from maximum height is solved using the equation \( 0 = v^2 - 2g(S+h) \), where \( S = \frac{v^2}{2g} \) is the maximum height reached. Solve for \( t_2 \) from the energy perspective or symmetry of motion: \( h + \frac{v^2}{2g} = \frac{1}{2}g(t_2)^2 \).

Step 5: Calculate Total Time

Sum \( t_1 \) and \( t_2 \) to obtain the total time from release to when the ball hits the ground. This involves solving \( t_2 \) and depends on \( h \). Re-evaluate to choose the square root option that applies.

Step 6: Compare Options and Choose Correct Answer

Evaluate the derived expression for total time \( T = \frac{v}{g} \left( 1 + \sqrt{1+\frac{2gh}{v^2}} \right) \) for consistency with the given options, leading to choice (a).

Key Concepts

Projectile Motion

Projectile motion is a fascinating topic in physics that involves the motion of an object thrown into the air, subject to only the force of gravity. When we talk about projectile motion, it includes any kind of motion that occurs in the vertical and horizontal planes simultaneously.
However, for the scenario in the exercise, the motion is purely vertical, making it a special case of projectile motion known as vertical motion. Here, a ball is tossed upward with a certain initial velocity, follows an upwards path until gravity halts its rise, and then returns back down.
This journey can be broken into two parts:

• The upward phase: The ball travels upward against gravity, slowing down until it stops completely at the peak of its flight.
• The downward phase: Once the peak is reached, the ball descends back down, accelerating due to gravity until it hits the ground.

Understanding this vertical projectile motion is crucial in predicting how long the object takes to reach each phase of its motion, which involves several calculations using kinematic equations.

Kinematic Equations

Kinematic equations are essential tools in solving problems related to the motion of objects. They help us describe how objects move, allowing us to calculate parameters like displacement, velocity, acceleration, and time.
For the ball thrown from the top of the tower, the relevant kinematic equation is:\[ s = ut + \frac{1}{2} at^2 \]where:

• \( s \) is the displacement, here being negative since the ball ends below the starting point.
• \( u \) is the initial velocity (\( v \)).
• \( a \) is the acceleration, equal to \( -g \) due to gravity's downward direction.
• \( t \) is the time duration the ball takes to travel.

By inputting the known values into this equation, the actions of the ball can be modeled mathematically.
This allows us to derive equations that predict the total time taken for the ball's journey from launch to when it strikes the ground, providing a deeper insight into the problem at hand.

Motion Under Gravity

The concept of motion under gravity is a key idea in physics. It describes how objects accelerate towards Earth due to the gravitational force. For objects like the ball in the problem, gravity is the only force acting upon it once it's in motion, pointing directly downwards.
Key aspects of motion under gravity include:

• Gravitational acceleration, typically denoted as \( g \), has a value of approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface.
• When an object is thrown upwards, it suffers a deceleration equivalent to \( -g \) until it momentarily stops at the zenith of its path.
• From the peak, the same gravitational force causes it to accelerate downwards, increasing its speed until it encounters another force (like impact with the ground).

Thus, understanding motion under gravity helps us to comprehensively analyze the descent of objects, utilizing derived equations to calculate things like time, speed, and final position quite efficiently. This is crucial for solving problems like calculating how long a ball launched upwards will take to fall back and hit the ground.