Problem 10 - Master Resource Book in JEE Main Physics

Question

If a stone is to hit at a point which is at a distance \(d\) away and at a height \(h\) above the point from where the stone starts, then what is the value of initial speed \(u\), if the stone is launched at an angle \(\theta\) ? (a) \(\frac{g}{\cos \theta} \sqrt{\frac{d}{2(d \tan \theta-h)}}\) (b) \(\frac{d}{\cos \theta} \sqrt{\frac{d}{2(d \tan \theta-h)}}\) (c) \(\sqrt{\frac{g d^{2}}{h \cos ^{2} \theta}}\) (d) \(\sqrt{\frac{g d^{2}}{(d-h)}}\)

Step-by-Step Solution

Answer

The correct answer is (c): \( \sqrt{\frac{g d^{2}}{h \cos ^{2} \theta}} \).

Step 1: Identify the given parameters

We know the following parameters: - Distance from the point: \( d \) - Height above from where the stone starts: \( h \) - Launch angle: \( \theta \) - Gravitational acceleration: \( g \) We need to find the initial speed \( u \).

Step 2: Use the equations of motion

For a projectile, the horizontal range \( R \) covered can be given by,\[R = \frac{u^2 \sin(2\theta)}{g}\]However, here we are considering height "\( h \)" and distance "\( d \)".

Step 3: Write horizontal and vertical displacement equations

The horizontal displacement \( d \) is given by:\[d = u \cos\theta \cdot t\]The vertical displacement for height \( h \) can be given by:\[h = u \sin\theta \cdot t - \frac{1}{2} gt^2\]

Step 4: Solve for time of flight from horizontal equation

From the equation \( d = u \cos\theta \cdot t \), we can solve for time \( t \):\[t = \frac{d}{u \cos\theta}\]

Step 5: Substitute time into vertical equation

Substitute \( t = \frac{d}{u \cos\theta} \) into the vertical motion equation:\[h = u \sin\theta \cdot \frac{d}{u \cos\theta} - \frac{1}{2} g \left( \frac{d^2}{u^2 \cos^2\theta} \right)\]

Step 6: Simplify and solve for initial speed \( u \)

Reorganize the previous equation:\[h = \frac{d \tan\theta}{g} u^2 \cos^2\theta - \frac{1}{2} \cdot \frac{g d^2}{u^2 \cos^2\theta}\]And solve for \( u \):\[u^2 = \frac{g d^2}{h \cos^2 \theta}\]Taking square root gives:\[u = \sqrt{\frac{g d^2}{h \cos^2 \theta}}\]

Step 7: Match with given options

The derived expression \( u = \sqrt{\frac{g d^2}{h \cos^2 \theta}} \) matches option (c).

Key Concepts

Initial Speed Calculation

When dealing with projectile motion, one of the first tasks is to determine the initial speed needed to reach a specific target. By knowing the initial velocity, angle of launch, and the positions involved, we can predict the object's trajectory.
In this exercise, we start by identifying the crucial parameters such as distance ( \( d \)), height ( \( h \)), launch angle ( \( \theta \)), and the gravitational acceleration ( \( g \)).

• Distance \( d \): This is the horizontal distance from the launch point to where the projectile lands.
• Height \( h \): This is the vertical distance above the launch point.
• Launch angle \( \theta \): The angle at which the projectile is fired.
• Gravitational acceleration \( g \): Usually given as \( 9.81 \, \text{m/s}^2 \).

To find the initial speed ( \( u \)), we solve the equations of motion that describe both horizontal and vertical movements. It's crucial to understand that both dimensions are interdependent and feed into each other to calculate the needed initial speed. By breaking down the motion into these parts, we can accurately compute the initial speed necessary to ensure the projectile reaches the desired new position, matching the criteria defined in the problem statement.

Projectile Range Equation

Understanding the projectile range equation is a pivotal part of solving problems about projectile motion. The range equation calculates the maximum horizontal distance a projectile travels, given its initial speed and the angle at which it is launched.
This important equation is expressed as: \[R = \frac{u^2 \sin(2\theta)}{g} \]
Where \( R \) is the range, \( u \) is the initial speed, \( \theta \) is the launch angle, and \( g \) is gravitational acceleration. However, in scenarios where height differences are involved, like having a target at height \( h \) instead of ground level, adjustments must be made.
For example:

• By substituting relevant terms for height and rearranging, the relationship can be adapted to reflect changes due to vertical displacement.
• This adjustment reflects the balance between the bias of speed lost to gravity and additional height gained.

Through these modifications, the horizontal range calculation informs our understanding of the velocity required to reach the target, balancing both horizontal distance and any vertical shift.

Vertical and Horizontal Displacement

Projectile motion involves the interplay of vertical and horizontal displacements. Understanding these components is essential to predicting the outcome of a projectile's path. Each aspect is guided by different, yet interrelated, principles.

For horizontal displacement: \[d = u \cos\theta \cdot t\]
The horizontal motion assumes constant velocity because gravity doesn't affect the horizontal component, meaning the horizontal displacement is a direct outcome of speed and time.
In vertical displacement: \[h = u \sin\theta \cdot t - \frac{1}{2} gt^2 \]
Vertical motion is more complex because it must take into account gravity, which affects how high or low the projectile will go over time.

• Gravity impacts the vertical component, dragging the projectile downwards, affecting both peak height and total travel time.
• The quadratic term \(-\frac{1}{2} gt^2\) shows gravity's influence over time, pulling the projectile back towards Earth.

To find the ideal launch conditions when both displacements are at play, one can solve these equations together, determining the time of flight first through horizontal displacement, then plugging back into the vertical to find the required velocity or range as needed for specific scenarios.