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Four bodies \(P, Q, R\) and \(S\) are projected with equal velocities having angles of projection \(15^{\circ}, 30^{\circ}, 45^{\circ}\) and \(60^{\circ}\) with the horizontal respectively. The body having shortest range is (a) \(P\) (b) \(Q\) (c) \(R\) (d) \(S\)

Short Answer

Expert verified
The body with the shortest range is \( P \).

Step by step solution

01

Identify the angle of projection for each body

The angles of projection are: \( P: 15^{\circ} \), \( Q: 30^{\circ} \), \( R: 45^{\circ} \), and \( S: 60^{\circ} \).
02

Apply the principle that the range is maximum at 45 degrees

The range is maximum for an angle of projection of \( 45^{\circ} \). This suggests that the range decreases for angles less than and greater than \( 45^{\circ} \). Therefore, the range for bodies for \( P \) and \( Q \) would be less than that for \( R \), and the range for \( S \) would also be less than that for \( R \).
03

Identify the body with the shortest range

Since \( 15^{\circ} \) is the smallest angle of projection, it will result in the shortest range. Therefore, body \( P \) will have the shortest range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Range of Projectile
In physics, understanding the range of a projectile is crucial when solving projectile motion problems. The range (R) of a projectile is the horizontal distance it travels before hitting the ground. It is calculated based on the initial speed, gravity, and the angle at which the projectile is launched relative to the horizontal plane.
The formula to calculate the range for a given angle of projection is:\[R = \frac{v^2 \sin(2\theta)}{g}\]where:
  • \(v\) is the initial velocity.
  • \(\theta\) is the angle of projection.
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2\) on Earth.
Thus, the greatest range occurs at an angle of 45 degrees. This is because \(\sin(90^{\circ})\) is equal to 1, which maximizes the value of the range formula. Understanding this concept allows students to predict how changes in the angle affect the distance a projectile travels.
Angle of Projection
The angle of projection plays a significant role in determining the range and trajectory of a projectile. Most students misunderstand how varying angles impact motion, as it not only affects distance but also the height and time a projectile stays in the air.
Projectile motion is symmetric, meaning that angles of "\(45^{\circ} - x\)" and "\(45^{\circ} + x\)" will yield the same range. Hence, both \(30^{\circ}\) and \(60^{\circ}\) result in the same range, though with different trajectories and maximum heights.
Understanding this symmetry helps in solving various physics problems where predicting the projectile's path is necessary. It's a useful tool for optimizing conditions in practical applications, such as sports, engineering, and various science fields.
Physics Problem Solving
Approaching physics problems systematically is key to finding the right solution. Several strategies can be employed to tackle such problems effectively:
  • **Break It Down:** Analyze the problem by identifying known quantities and what needs to be solved.
  • **Apply Physics Principles:** Use relevant formulas and concepts, such as kinematics equations for projectile motion.
  • **Draw Diagrams:** Visual aids can help understand complex trajectories and interactions.
  • **Check Your Work:** Verify calculations and ensure results make physical sense in the context of the problem.
By adopting these practices, students can improve their problem-solving skills and develop a deeper understanding of physics concepts, leading to more accurate and reliable solutions.

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Most popular questions from this chapter

A smooth table is placed horizontally and an ideal spring of spring constant \(k=1000 \mathrm{~N} / \mathrm{m}\) and unextended length of \(0.5 \mathrm{~m}\) has one end fixed to its centre. The other end is attached to a mass of \(5 \mathrm{~kg}\) which is moving in a circle with constant speed \(20 \mathrm{~m} / \mathrm{s}\). Then the tension in the spring and the extension of this spring beyond its normal length are (a) \(500 \mathrm{~N}, 0.5 \mathrm{~m}\) (b) \(600 \mathrm{~N}, 0.6 \mathrm{~m}\) (c) \(700 \mathrm{~N}, 0.7 \mathrm{~m}\) (d) \(800 \mathrm{~N}, 0.8 \mathrm{~m}\)

A particle moving along the circular path with a speed \(v\) and its speed increases by ' \(g\) ' in one second. If the radius of the circular path be \(r\), then the net acceleration of the particle is (a) \(\frac{v^{2}}{r}+g\) (b) \(\frac{v^{2}}{r^{2}}+g^{2}\) (c) \(\left[\frac{v^{4}}{r^{2}}+g^{2}\right]^{\frac{1}{2}}\) (d) \(\left[\frac{v^{2}}{r}+g\right]^{\frac{1}{2}}\)

With what angular velocity should a \(20 \mathrm{~m}\) long cord be rotated such that tension in it, while reaching the highest point, is zero (a) \(0.5 \mathrm{rad} / \mathrm{sec}\) (b) \(0.2 \mathrm{rad} / \mathrm{sec}\) (c) \(7.5 \mathrm{rad} / \mathrm{sec}\) (d) \(0.7 \mathrm{rad} / \mathrm{sec}\)

A body is revolving with a uniform speed \(v\) in a circle of radius \(r\). The angular acceleration of the body is (a) \(\frac{v}{r}\) (b) Zero (c) \(\frac{v^{2}}{r}\) along the radius and towards the centre (d) \(\frac{v^{2}}{r}\) along the radius and away from the centre

If the equation for the displacement of a particle moving on a circular path is given by \((\theta)=2 t^{3}+0.5\), where \(\theta\) is in radians and \(t\) in seconds, then the angular velocity of the particle after \(2 \mathrm{sec}\) from its start is [AIIMS 1998] (a) \(8 \mathrm{rad} / \mathrm{sec}\) (b) \(12 \mathrm{rad} / \mathrm{sec}\) (c) \(24 \mathrm{rad} / \mathrm{sec}\) (d) \(36 \mathrm{rad} / \mathrm{sec}\)

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