Problem 34 - Master Resource Book in JEE Main Physics

Question

The velocity of projection of an oblique projectile is \((6 \hat{\mathbf{i}}+\mathbf{8} \hat{\mathbf{j}}) \mathrm{ms}^{-1}\). The horizontal range of the projectile is (a) \(4.9 \mathrm{~m}\) (b) \(9.6 \mathrm{~m}\) (c) \(19.6 \mathrm{~m}\) (d) \(14 \mathrm{~m}\)

Step-by-Step Solution

Answer

The correct option is (b) 9.6 meters.

Step 1: Identify Components of Initial Velocity

The given velocity vector is \(\mathbf{v} = (6 \hat{\mathbf{i}} + 8 \hat{\mathbf{j}}) \mathrm{ms}^{-1}\). From this, identify the horizontal component \(v_x\) as 6 m/s and the vertical component \(v_y\) as 8 m/s.

Step 2: Calculate Time of Flight

The time of flight \(T\) for a projectile is given by the formula \(T = \frac{2v_y}{g}\), where \(g = 9.8 \mathrm{ms}^{-2}\) is the acceleration due to gravity. Substituting the value of \(v_y\), we have:\[ T = \frac{2 \times 8}{9.8} \approx 1.63 \text{ seconds}. \]

Step 3: Calculate Horizontal Range

The horizontal range \(R\) is given by the formula \(R = v_x \times T\). Using the values from previous steps, we have:\[ R = 6 \times 1.63 \approx 9.78 \text{ meters}. \]

Step 4: Compare with Given Options

The calculated horizontal range is approximately 9.78 meters. Compare with the closest given option which is 9.6 meters (option b).

Key Concepts

Velocity Components

In projectile motion, understanding the components of velocity is crucial for solving related problems. The initial velocity of a projectile can often be broken down into two perpendicular components: horizontal and vertical. When given a velocity vector like \(\mathbf{v} = (6 \hat{\mathbf{i}} + 8 \hat{\mathbf{j}}) \, \mathrm{ms}^{-1}\), it means we have:

• Horizontal component (\(v_x\)) represented by \(6 \,\mathrm{m/s}\): This shows how fast the projectile moves sideways.
• Vertical component (\(v_y\)) represented by \(8 \,\mathrm{m/s}\): This indicates how quickly it rises or falls.

Breaking the velocity into these components allows us to analyze the motion separately in horizontal and vertical planes. This is particularly useful because the forces acting on the projectile, like gravity, affect these planes differently.

Time of Flight

The time of flight is a key factor in projectile motion, representing the total time the projectile remains in the air. To calculate it for an oblique projectile, the formula used is \(T = \frac{2v_y}{g}\), where \(v_y\) is the initial vertical velocity and \(g\) is the acceleration due to gravity, approximately \(9.8 \,\mathrm{ms}^{-2}\).
For our example, the vertical component is \(8 \,\mathrm{m/s}\). Plugging these values into the formula gives:\[ T = \frac{2 \times 8}{9.8} \approx 1.63 \,\text{seconds} \]This tells us that the projectile will take roughly 1.63 seconds to reach back to the ground once launched. As gravity predominantly affects the vertical component, it primarily influences the time the projectile stays airborne.

Horizontal Range

The horizontal range of a projectile refers to the distance it travels along the horizontal axis during its flight. Calculating this involves multiplying the horizontal velocity component with the time of flight. The formula is:\[ R = v_x \times T \]Where \(v_x\) is the horizontal component of velocity, which is \(6 \,\mathrm{m/s}\) for our scenario, and \(T\) is the time of flight, \(1.63 \,\text{seconds}\).

• Substituting these values into the formula gives:\[ R = 6 \times 1.63 \approx 9.78 \,\text{meters} \]

This is the theoretical distance the projectile would cover along the ground, assuming no air resistance. Essentially, this calculation shows how far across a field (or any horizontal surface) the projectile will land from its starting point.