Problem 66 - Master Resource Book in JEE Main Physics

Question

A projectile is fired at an angle of \(30^{\circ}\) to the horizontal such that the vertical component of its initial velocity is \(80 \mathrm{~ms}^{-1}\), Its time of flight is \(T\). Its velocity at \(t=\frac{T}{4}\) has a magnitude of nearly (a) \(200 \mathrm{~ms}^{-1}\) (b) \(300 \mathrm{~ms}^{-1}\) (c) \(140 \mathrm{~ms}^{-1}\) (d) \(100 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

c) 140 ms⁻¹

Step 1: Identify Vertical Component

We are given that the vertical component of the projectile's initial velocity is \(80 \text{ m/s}\). This can be calculated using the initial velocity \(v_0\) and the angle \(30^{\circ}\) as \(v_{0y} = v_0 \sin 30^{\circ}\).

Step 2: Calculate Initial Velocity

Since \(\sin 30^{\circ} = \frac{1}{2}\), we have \(v_0 \times \frac{1}{2} = 80 \text{ m/s}\). Solving this gives \(v_0 = 160 \text{ m/s}\).

Step 3: Determine Horizontal Component

The horizontal component of the initial velocity can be calculated as \(v_{0x} = v_0 \cos 30^{\circ}\). Using \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\), we get \(v_{0x} = 160 \times \frac{\sqrt{3}}{2} = 80\sqrt{3} \text{ m/s}\).

Step 4: Calculate Velocity Components at t = T/4

At \(t = \frac{T}{4}\), the horizontal component of velocity remains unchanged, \(v_{x} = 80\sqrt{3} \text{ m/s}\). The vertical component at this time can be calculated using \(v_y = v_{0y} - g\left(\frac{T}{4}\right)\). First, we determine the time of flight \(T = \frac{2v_{0y}}{g} = \frac{2 \times 80}{10} = 16 \text{ s}\). So, \(v_y = 80 - 10 \times \frac{16}{4} = 80 - 40 = 40 \text{ m/s}\).

Step 5: Calculate Resultant Velocity Magnitude

The magnitude of velocity at \(t = \frac{T}{4}\) can be found using Pythagorean theorem: \(v = \sqrt{v_x^2 + v_y^2}\). Substituting values gives \(v = \sqrt{(80\sqrt{3})^2 + (40)^2}\). This simplifies to \(v = \sqrt{19200 + 1600} = \sqrt{20800} \approx 144.2 \text{ m/s}\).

Step 6: Determine the Closest Answer Option

The calculated magnitude \(144.2 \text{ m/s}\) is closest to option \(c) \; 140 \text{ m/s}\).

Key Concepts

Initial Velocity Calculation

In projectile motion, determining the initial velocity is crucial to understanding how the projectile will move. The initial velocity (\(v_0\)) is split into horizontal and vertical components. Here, we begin with the vertical component, which is usually given or can be calculated from the problem statement. In this exercise, the vertical velocity component is provided as \(80 \text{ m/s}\) and occurs at a \(30^{\circ}\) angle to the horizontal. Thus, we use the formula: \(v_{0y} = v_0 \sin(30^{\circ})\).

• Since \(\sin(30^{\circ}) = \frac{1}{2}\), we know \(v_0 \times \frac{1}{2} = 80 \text{ m/s}\).
• This equation can be rearranged to solve for the initial velocity: \(v_0 = 160 \text{ m/s}\).

This provides us the total initial velocity before it is split into its components, which is integral for subsequent calculations.

Time of Flight

The time of flight is the total duration the projectile spends in the air from launch to landing. For symmetric projectile motion on a flat surface, the formula is \(T = \frac{2v_{0y}}{g}\), where \(g\) is the acceleration due to gravity. Here, with \(v_{0y} = 80 \text{ m/s}\) and \(g = 10 \text{ m/s}^2\), we calculate:

• \(T = \frac{2 \times 80}{10} = 16 \text{ s}\).

This shows the projectile will remain airborne for 16 seconds, providing the foundation for further investigation into other aspects of its motion.

Velocity Components

Breaking down the initial velocity into its components helps analyze the projectile's trajectory. These components remain constant or vary differently—a critical insight into projectile motion:

Vertical Component

Initially \(80 \text{ m/s}\), the vertical component changes over time due to gravity. At \(t = \frac{T}{4}\), it is determined by:
\(v_y = v_{0y} - g\left(\frac{T}{4}\right)\).

• \(v_y = 80 - 10 \times 4 = 40 \text{ m/s}\).

Horizontal Component

The horizontal component \(v_{0x}\) uses \(\cos(30^{\circ})\):

• \(v_{0x} = 160 \times \frac{\sqrt{3}}{2} = 80\sqrt{3} \text{ m/s}\).
• This remains unchanged throughout the flight since there's no horizontal acceleration.

By keeping these components in mind, we build toward the resultant velocity needed for understanding the projectile's path.

Resultant Velocity

When analyzing a projectile, combining both horizontal and vertical components gives the resultant velocity. At any specific time, like \(t = \frac{T}{4}\), we employ the Pythagorean theorem:
\(v = \sqrt{v_x^2 + v_y^2}\).

• Insert the known values: \(v = \sqrt{(80\sqrt{3})^2 + (40)^2}\).
• Calculate, leading to: \(v = \sqrt{19200 + 1600} = \sqrt{20800} \approx 144.2 \text{ m/s}\).

This magnitude provides concrete data on the projectile's velocity at the specified time point, pivotal for tasks like pinpointing trajectory or impact speeds.

Angle of Projection

The angle of projection, a fundamental ingredient in projectile dynamics, influences both range and peak height. Set at \(\theta = 30^{\circ}\) in this case, it defines the initial division between vertical and horizontal force (by using trigonometric functions like \(\sin\) and \(\cos\)). The angle affects how far and high the projectile will go:

• A smaller angle results in a lower, longer trajectory while a steeper angle leads to a higher, shorter path.

Properly understanding this relationship equips one to make informed predictions or modifications to projectile motion outcomes in real-life applications, like sports or engineering.