Question

Angle of projection of a projectile is changed, keeping initial velocity constant. Find the rate of change of maximum height. Range of the projectile is \(\mathrm{R}\). (A) \((\mathrm{R} / 4)\) (B) \((\mathrm{R} / 3)\) (C) \((\mathrm{R} / 2)\) (D) \(\mathrm{R}\)

Short answer

The rate of change of maximum height with respect to the angle of projection, while keeping the initial velocity constant, is equal to the range of the projectile. Therefore, the short answer is: \(R\)

Step-by-step solution

Recall the formula for range and maximum height of a projectile

The formulas we need are: Range: \[R = \frac{v_0^2 \sin(2\theta)}{g}\] Maximum Height: \[H = \frac{v_0^2 \sin^2(\theta)}{2g}\] where \(v_0\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

Differentiate the maximum height formula with respect to the angle of projection

We want to find the rate of change of maximum height with respect to the angle, which means we need to differentiate the height formula with respect to \(\theta\). Using the chain rule, we have: \[\frac{dH}{d\theta} = \frac{v_0^2}{2g} \cdot 2 \sin(\theta) \cos(\theta)\] Simplifying, we get: \[\frac{dH}{d\theta} = \frac{v_0^2 \sin(2\theta)}{g}\]

Rewrite the expression for the rate of change in terms of range

The expression for the rate of change of maximum height is the same as the expression for the range, except for a constant factor. Hence, we can write: \[\frac{dH}{d\theta} = R\] where \(R\) is the range of the projectile.

Compare the rate of change with the given options

Now, we just have to compare this result with the given options: (A) 𝑅/4 (B) 𝑅/3 (C) 𝑅/2 (D) 𝑅 The rate of change of maximum height is equal to the range of the projectile, so the correct answer is: (D) 𝑅

Key concepts

Range of Projectile

In physics, the range of a projectile is the horizontal distance it travels before hitting the ground. It is an essential concept in projectile motion, helping us understand how various factors influence a projectile's trajectory. When we talk about the range, we're interested in the distance from the launch point to the point where the projectile returns to the same vertical level. The formula to calculate this distance is: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] Where:

• \( R \) represents the range of the projectile.
• \( v_0 \) is the initial velocity, or the speed at which the projectile is launched.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \ m/s^2 \) on Earth.

The range depends on both the initial speed and the launch angle. Specifically, for the maximum range on level ground, the optimal angle is \( 45^{\circ} \). This is because the sine function reaches its maximum value of 1 when \( 2\theta = 90^{\circ} \). Understanding the range helps in various applications, such as sports, military, and engineering, wherever projectile motion is relevant.

Maximum Height of Projectile

The maximum height of a projectile refers to the peak vertical position that it reaches in its trajectory. It's a crucial parameter when analyzing the motion of projectiles, as it tells us how high the projectile will go, which can be important for applications like sports or engineering.The equation for calculating the maximum height is:\[ H = \frac{v_0^2 \sin^2(\theta)}{2g} \]Where:

• \( H \) is the maximum height of the projectile.
• \( v_0 \) is the initial velocity.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity.

This formula shows that the maximum height depends on the square of the sine of the angle of projection, emphasizing the need for sufficient vertical velocity to achieve a higher altitude. The initial speed also plays a critical role; the greater the speed, the higher the maximum height, as long as the vertical component of the velocity is significant.Understanding the maximum height provides insights into the potential coverage and safety considerations in a variety of situations.

Rate of Change in Maximum Height

Analyzing the rate of change in the maximum height of a projectile reveals how sensitive the peak vertical position is to changes in the launch angle. It helps us see how minute adjustments to the angle can significantly affect the height achieved.To find the rate of change of maximum height with respect to the angle of projection, we differentiate the maximum height formula concerning \( \theta \):\[ \frac{dH}{d\theta} = \frac{v_0^2}{2g} \cdot 2 \sin(\theta) \cos(\theta) \]Simplified using trigonometric identities, it becomes:\[ \frac{dH}{d\theta} = \frac{v_0^2 \sin(2\theta)}{g} \]Interestingly, this expression matches the formula for the range \( R \), indicating that the rate of change of maximum height is directly proportional to the projectile's range.This direct relationship implies that as the angle is adjusted, the change in maximum height corresponds consistently with the change in range, given the initial conditions stay constant. This insight is particularly valuable when designing projectiles for specific ranges or maximum heights in various fields.