Problem 63 - Master Resource Book in JEE Main Physics

Question

An aircraft, diving at an angle of \(53.0^{\circ}\) with the vertical releases a projectile at an altitude of \(730 \mathrm{~m}\). The projectile hits the ground \(5.00 \mathrm{~s}\) after being released. What is the speed of the aircraft? (a) \(282 \mathrm{~ms}^{-1}\) (b) \(202 \mathrm{~ms}^{-1}\) (c) \(182 \mathrm{~ms}^{-1}\) (d) \(102 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The speed of the aircraft is 202 m/s (option b).

Step 1: Analyze the Vertical Motion

We know that the projectile is released at a vertical height of 730 m and hits the ground 5 seconds later. We use the equation for vertical motion under gravity: \[ y = v_{0y}t + \frac{1}{2}gt^2 \] where \( y = -730 \) m (since it hits the ground), \( g = -9.8 \) m/s\(^2\) (acceleration due to gravity, taken as negative), and \( t = 5 \) s. The initial vertical velocity \( v_{0y} = v_0 \cos(53^{\circ}) \). Substitute these into the equation to find \( v_0 \).

Step 2: Simplify the Vertical Motion Equation

Substitute values from Step 1 into the vertical motion equation:\[ -730 = v_0 \cos(53^{\circ}) \, 5 + \frac{1}{2}(-9.8)(5)^2 \]Simplify the equation:\[ -730 = 5v_0 \cos(53^{\circ}) - 122.5 \]\[ -730 + 122.5 = 5v_0 \cos(53^{\circ}) \]\[ -607.5 = 5v_0 \cos(53^{\circ}) \]

Step 3: Calculate the Initial Speed of the Aircraft

To solve for \( v_0 \), calculate \( \cos(53^{\circ}) \), which approximately equals 0.6018.Now solve the equation:\[ -607.5 = 5v_0 (0.6018) \]\[ v_0 = \frac{-607.5}{5 \times 0.6018} \]\[ v_0 = \frac{-607.5}{3.009} \]\[ v_0 \approx 202 \, \mathrm{ms}^{-1} \]

Step 4: Conclude the Speed of the Aircraft

The calculated speed of the aircraft \( v_0 \) is approximately 202 m/s. Verify this aligns with the given options. Thus, the answer is (b) 202 m/s.

Key Concepts

Vertical Motion Equations

When analyzing projectile motion, particularly in the vertical direction, we use the vertical motion equations. These equations help us understand how objects move under the influence of gravity. The basic equation is: \[ y = v_{0y}t + \frac{1}{2}gt^2 \] where:

• \( y \) is the vertical displacement, which is negative if the object falls down.
• \( v_{0y} \) is the initial vertical velocity.
• \( g \) represents the acceleration due to gravity, approximately \(-9.8\, \mathrm{m/s}^2\) on Earth.
• \( t \) is the time the object is in motion.

In our exercise, the projectile is released from an altitude of 730 m. It travels vertically under the influence of gravity for 5 seconds. By substituting these values into our equation, we can determine the initial speed of the projectile's release when broken down into vertical components.

Trigonometric Functions in Physics

Trigonometric functions, like cosine and sine, are essential in physics to resolve vectors into their components. In projectile motion, these functions help us convert the initial speed of an object into its horizontal and vertical components. When dealing with angles, such as the aircraft diving at \(53.0^{\circ}\), we use:

• \( v_{0y} = v_0 \cos(\theta) \), for the vertical component.
• \( v_{0x} = v_0 \sin(\theta) \), for the horizontal component.

These equations use: - \( \theta \) for the angle of the dive (53.0° in this problem). - \( v_0 \) for the initial speed. For our problem, the vertical component \( v_{0y} \) was found using cosine, critical to solving the initial speed of the aircraft. Calculating \( \cos(53^{\circ}) \) as approximately 0.6018 was a crucial step to finding the actual speed of the projectile.

Kinematics of Free Fall

In free fall, objects are only influenced by gravity, meaning they experience constant acceleration downward. This makes solving motion straightforward, as the acceleration is always \(-9.8\, \mathrm{m/s^2}\) on Earth. Several equations simplify the analysis of free fall, such as: \[ y = v_{0}t - \frac{1}{2}gt^2 \] This form accounts for when initial vertical velocity might be zero or non-zero, and it can be rearranged to solve for different variables, knowing the constant acceleration. In the context of the exercise, we treated the projectile as entering free fall continuously upon release. Understanding free fall helps comprehend how it would hit the ground in 5 seconds, given an initial velocity derived from the diving aircraft. The kinematics of free fall underlines why the time of fall was crucial in back-calculating the initial speed, ensuring we factored in the constant gravitational pull correctly.