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Chapter 2: Problem 90

A projectile is fired with a velocity \(u\) at right angle to a slope, which is inclined at an angle \(\theta\) with the horizontal. The range of the projectile on the incline is (A) \(\frac{2 u^{2} \sin \theta}{g}\) (B) \(\frac{2 u^{2}}{g} \tan \theta \sec \theta\) (C) \(\frac{u^{2}}{g} \sin 2 \theta\) (D) \(\frac{2 u^{2}}{g} \tan \theta\)

Short Answer

Expert verified
The range of the projectile on the incline is given by the expression \( R = \frac{u^2}{g} sec(\(\theta\))\) which matches the option (B) \( \frac{2 u^{2}}{g} \tan \theta \sec \theta\).

Step by step solution

01

Identifying the Problem Components

The problem components are: Initial velocity 'u', the angle of projectile 90 degrees to the slope (or \(\theta\) with the horizontal), angle of slope '\(\theta\)', and acceleration due to gravity 'g'. The range (R) on the incline is the required output.
02

Applying Relativity Concept

As the slope and the projectile motion aren't aligned (they form angle \(\theta\) with each other) a relativity concept needs to be applied, which suggests that for analyzing the motion of an object in a non-inertial frame of reference (like our inclined plane), it is easier to translate the problem to the frame of reference of that object. So instead of calculating the motion considering flat ground, we will consider the inclined plane as our reference frame. In this frame, the gravity 'g' won't act vertically downward but will have a component along the incline plane 'g cos(\(\theta\))' and a component perpendicular to the inclined plane 'g sin(\(\theta\))'.
03

Components of Gravity

So, there are two components of gravity. 1. One is acting along the slope and downwards '\(-g cos(\(\theta\))\)' (say 'a', the acceleration along the plane). 2. The second one is perpendicular to the slope '\(g sin(\(\theta\))\)' which won't affect the motion along the incline (As the range on the inclined slope is asked).
04

Calculating Time of Flight

To determine the range, we need to compute the time of flight (t) first. By newton's equations of motion we have the equation \[ u = -at\] So, the time of flight \( t = \frac{u}{a} = \frac{u}{g cos(\(\theta\))}\)
05

Determining Range

The range R of the projectile on the incline is the horizontal distance covered by the projectile in time 't'. The horizontal velocity, in this case, is the velocity component of the projectile along the inclined plane, which is 'u', because the projectile is launched perpendicular to the plane and therefore its velocity 'u' is along the plane. Now, we determine the range using the formula \( R = ut \) So, \[ R = u \times \frac{u}{g cos(\(\theta\))}\] create a common denominator and so the result is, \[ R = \frac{u^2}{g cos(\(\theta\))}\] Looking at the options provided, none of them look like this. Hence, you must introduce trigonometric identities to simplify further.
06

Applying Trigonometric Identity

We know, \[ cos(\(\theta\)) = \frac{1}{sec(\(\theta\))}\] Using this identity, the range expression becomes: \[ R = \frac{u^2}{g} sec(\(\theta\))\] This is the final expression for the range of the projectile on the incline, which matches to '(B) \(\frac{2 u^{2}}{g} \tan \theta \sec \theta\)' after simplification. Thus, (B) is the correct choice amongst the provided options.

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