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Chapter 5: Problem 4

The range \(R\) of projectile is same when its maximum heights are \(h_{1}\) and \(h_{2} .\) What is the relation between \(R, h_{1}\), and \(h_{2} ?\) (1) \(R=\sqrt{h_{1} h_{2}}\) (2) \(R=\sqrt{2 h_{1} h_{2}}\) (3) \(R=2 \sqrt{h_{1} h_{2}}\) (4) \(R=4 \sqrt{h_{1} h_{2}}\)

Short Answer

Expert verified
The relation is \(R = 4 \sqrt{h_1 h_2}\).

Step by step solution

01

Understand the Problem

We need to find the relation between the range \(R\) of a projectile and its maximum heights \(h_1\) and \(h_2\). The projectile has the same range for two different angles, resulting in different maximum heights.
02

Range and Maximum Height Formulas

The range \(R\) of a projectile is given by \(R = \frac{v^2 \sin(2\theta)}{g}\), where \(v\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity. The maximum height \(h\) for a given angle is: \(h = \frac{v^2 \sin^2(\theta)}{2g}\).
03

Relationship Between Height and Angle

Using the formulas for maximum height, we have for \(h_1\): \(h_1 = \frac{v^2 \sin^2(\theta_1)}{2g}\), and for \(h_2\): \(h_2 = \frac{v^2 \sin^2(\theta_2)}{2g}\), where \(\theta_1\) and \(\theta_2\) are the angles of projection.
04

Relate the Two Maximum Heights to Range

Since the range is the same for both \(\theta_1\) and \(\theta_2\), we can set the range formulas equal: \(\frac{v^2 \sin(2\theta_1)}{g} = \frac{v^2 \sin(2\theta_2)}{g}\). As \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\), the angles \(\theta_1\) and \(\theta_2\) satisfy: \(\sin(2\theta_1) = \sin(2\theta_2)\).
05

Derive the Relation Between \(R\) and \(h_1, h_2\)

From Step 3, substituting the expressions of \(\sin(\theta_1)\) and \(\sin(\theta_2)\) derived from their respective heights into \(\sin(2\theta)\), and multiplying, we get:\[\sin(2\theta_1) \sin(2\theta_2) = 4 \sin(\theta_1) \cos(\theta_1) \sin(\theta_2) \cos(\theta_2) = 4 \times \frac{\sin^2(\theta_1)}{2}\times \frac{\sin^2(\theta_2)}{2}\]Therefore, \(R = 4 \sqrt{h_1 h_2}\), which matches with option (4).
06

Final Answer

The correct relation between \(R\), \(h_1\), and \(h_2\) is \(R = 4 \sqrt{h_1 h_2}\).

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Key Concepts

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

Range of Projectile
The range of a projectile is a crucial concept in projectile motion. It is the horizontal distance a projectile travels while in motion. Understanding range helps in determining how far a projectile will land from its starting point. The formula for range is given by the equation: \[ R = \frac{v^2 \sin(2\theta)}{g} \] where
  • \( R \) is the range,
  • \( v \) is the initial velocity,
  • \( \theta \) is the angle of projection,
  • \( g \) is the acceleration due to gravity.
The range is highly dependent on the angle of projection. For a fixed initial velocity, the maximum range is achieved when the angle of projection is 45 degrees. However, what's interesting is that for any angle \( \theta_1 \), there is a complementary angle \( \theta_2 \) (where \( \theta_1 + \theta_2 = 90 \) degrees) that gives the same range due to the symmetry of the sine function in the range formula.
Maximum Height
The maximum height of a projectile is the highest vertical point it reaches during its flight. Calculating the maximum height gives insights into how high the projectile goes before descending back down. The formula for maximum height is: \[ h = \frac{v^2 \sin^2(\theta)}{2g} \] where
  • \( h \) is the maximum height,
  • \( v \) is the initial velocity,
  • \( \theta \) is the angle of projection,
  • \( g \) is the acceleration due to gravity.
Maximum height is directly influenced by the angle and speed at which the projectile is launched. A greater angle generally means a higher arc, but the height also depends on the sine component, which attains its highest value at 90 degrees. Thus, to reach the greatest height for a given initial velocity, the projectile must be launched vertically. However, in combination with a horizontal distance, or range, this factor is part of the balance that determines the projectile's trajectory.
Angle of Projection
The angle of projection plays a critical role in the trajectory of a projectile. It determines both the maximum height and range. The angle of projection \( \theta \) is the angle at which an object is launched above the horizontal axis. It helps in defining the path of the projectile and can be used to infer certain behaviors. When a projectile is launched, angles interact with the initial velocity to form the specific parabola path of a trajectory.
  • At angles less than 45 degrees, the range is favored more than the height.
  • At exactly 45 degrees, both range and maximum height are optimized for synergy.
  • At angles greater than 45 degrees, the projectile will reach higher points but cover less distance horizontally.
Ultimately, the angle of projection decides if the motion will favor height or range. In scenarios like sports or combat, optimizing the angle can lead to an effective trajectory, maximizing either distance or peak depending on the goal.

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Most popular questions from this chapter

Two balls \(A\) and \(B\) are thrown with speeds \(u\) and \(u\) ? respectively. Both the balls cover the same horizont distance before returning to the plane of projection. If th angle of projection of ball \(B\) is \(15^{\circ}\) with the horizontal, the the angle of projection of \(A\) is (1) \(\sin ^{-1}\left(\frac{1}{8}\right)\) (2) \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{8}\right)\) (3) \(\frac{1}{3} \sin ^{-1}\left(\frac{1}{8}\right)\) (4) \(\frac{1}{4} \sin ^{-1}\left(\frac{1}{8}\right)\)

Two ports, \(A\) and \(B\), on a North-South line are separated by a \(\pi_{\text {le }}\) of width \(D\). The river flows east with speed \(u\). A boat crosses the river starting from port \(A .\) The speed of the boat relative to the river is \(v\). Assume \(v=2 u\). Suppose the boat wants to cross the river from \(A\) to the other side in the shortest possible time. Then what should be the direction of the velocity of boat relative to river? (1) \(30^{\circ}\) west of north (2) \(30^{\circ}\) east of north (3) \(60^{\circ}\) west of north (4) along north

A shot is fired from a point at a distance of \(200 \mathrm{~m}\) from the foot of a tower \(100 \mathrm{~m}\) high so that it just passes over i horizontally. The direction of shot with horizontal is (1) \(30^{\circ}\) (2) \(45^{\circ}\) (3) \(60^{\circ}\) (4) \(70^{\circ}\)

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. The distance of the point from point \(O\) where the package lands is (1) \(80 \mathrm{~m}\) (2) \(100 \mathrm{~m}\) (3) \(200 \mathrm{~m}\) (4) \(160 \mathrm{~m}\)

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate The velocity with which particle strikes the plane \(O B\), (1) \(15 \mathrm{~ms}^{-1}\) (2) \(30 \mathrm{~ms}^{\prime}\) (3) \(20 \mathrm{~ms}^{-1}\) (4) \(10 \mathrm{~ms}^{-1}\)
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