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Chapter 4: Problem 26

When a projectile is projected at a certain angle with the horizontal, its horizontal range is \(R\) and time of flight is \(T_{1}\). When the same projectile is throwing with the same speed at some other angle with the horizontal, its horizontal range is \(R\) and time of flight is \(T_{2}\). The produet of \(T_{1}\) and \(T_{2}\) is (a) \(\frac{R}{g}\) (b) \(\frac{2 R}{g}\) (c) \(\frac{3 R}{g}\) (d) \(\frac{4 R}{g}\)

Short Answer

Expert verified
The product \( T_1 \times T_2 \) is \( \frac{2R}{g} \), option (b).

Step by step solution

01

Recall the formula for time of flight

The time of flight for a projectile launched at an angle \( \theta \) with an initial velocity \( v \) is given by \( T = \frac{2v \sin \theta}{g} \), where \( g \) is the acceleration due to gravity.
02

Relate time of flight to horizontal range

The horizontal range \( R \) for a projectile launched with speed \( v \) at an angle \( \theta \) can be found using \( R = \frac{v^2 \sin(2\theta)}{g} \).
03

Express time of flights \( T_1 \) and \( T_2 \)

Given that the projectile has the same horizontal range \( R \) for two angles, denote the time of flights as \( T_1 = \frac{2v \sin \theta_1}{g} \) and \( T_2 = \frac{2v \sin \theta_2}{g} \).
04

Use trigonometric identity for range

For the same horizontal range, we have that \( \sin(2\theta_1) = \sin(2\theta_2) \). Using the identity \( \sin(A) = \sin(180^\circ - A) \), it follows that \( \theta_2 = 90^\circ - \theta_1 \).
05

Calculate \( \sin \theta_1 \) and \( \sin(90^\circ - \theta_1) \)

Since \( \theta_2 = 90^\circ - \theta_1 \), then \( \sin \theta_2 = \cos \theta_1 \). Hence, \( T_1 = \frac{2v \sin \theta_1}{g} \) and \( T_2 = \frac{2v \cos \theta_1}{g} \).
06

Compute \( T_1 \times T_2 \)

Multiply the expressions for \( T_1 \) and \( T_2 \): \[ T_1 \times T_2 = \left(\frac{2v \sin \theta_1}{g}\right) \times \left(\frac{2v \cos \theta_1}{g}\right) \] \[ = \frac{4v^2 (\sin \theta_1 \cos \theta_1)}{g^2} \] \[ = \frac{2v^2 \sin(2\theta_1)}{g^2} \] Using the range formula \( R = \frac{v^2 \sin(2\theta_1)}{g} \), we get: \[ \frac{2R}{g^2} \] Finally, it can be simplified to \( \frac{2R}{g} \).
07

Choose the correct option

The product \( T_1 \times T_2 = \frac{2R}{g} \), which corresponds to option (b).

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Key Concepts

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

Horizontal Range
The horizontal range is an essential concept in projectile motion, describing how far a projectile travels along the horizontal plane. To compute the horizontal range, we use the equation:\[ R = \frac{v^2 \sin(2\theta)}{g} \]where:
  • \( v \) is the initial velocity of the projectile.
  • \( \theta \) is the angle of launch with respect to the horizontal.
  • \( g \) is the acceleration due to gravity.
This formula indicates that the range depends on both the speed and angle of projection. It is maximized when \( 2\theta = 90^\circ \), or \( \theta = 45^\circ \). Keep in mind that this equation assumes no air resistance and launching from a level surface.
Time of Flight
Time of flight refers to the total time a projectile is in motion from launch until it hits the ground. Calculating the time of flight is critical for understanding how far and how long a projectile will be airborne. For a projectile launched at an angle \( \theta \) with initial velocity \( v \), the time of flight \( T \) is given by:\[ T = \frac{2v \sin \theta}{g} \]Key points include:
  • The time of flight increases with both higher initial velocities and more significant launch angles (up to 90掳).
  • It reaches a maximum when the projectile is launched vertically (\( \theta = 90^\circ \)).
  • This parameter is essential when solving real-world problems involving projectile motion, like determining where an object will land.
Understanding how time of flight interrelates with other elements like horizontal range is vital for solving complex problems in physics.
Trigonometric Identities
Trigonometric identities help us simplify expressions and solve problems involving angles and lengths in projectile motion. The key identity related to the range of a projectile is:\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]This identity is crucial because:
  • It allows us to convert between expressions involving \( \sin \) and \( \cos \).
  • In the context of horizontal range, it explains why two different angles can yield the same range, given that \( \sin(2\theta_1) = \sin(2\theta_2) \).
Moreover, the identity \( \sin(90^\circ - \theta) = \cos(\theta) \) is often used to resolve problems where angles are complementary, further illustrating the symmetry in projectile paths.
Acceleration due to Gravity
Acceleration due to gravity, denoted as \( g \), is a constant force that pulls objects towards the Earth. In projectile motion, this force plays a crucial role as it influences both the vertical motion of the projectile and its trajectory:
  • The standard value is approximately \( 9.81 \text{ m/s}^2 \), depending slightly on geographic location.
  • It is the sole vertical force acting on a projectile in the absence of air resistance.
  • In equations, \( g \) affects calculations such as time of flight and maximum height.
Understanding \( g \) helps us predict how a projectile will behave鈥攆rom the moment it's launched to when it lands. The consistent value of gravity allows for straightforward calculations in theoretical contexts where other forces are negligible. Knowing \( g \) is vital in learning and applying the laws of motion.

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Most popular questions from this chapter

The velocity of projection of an oblique projectile is \((6 \hat{\mathbf{i}}+\mathbf{8} \hat{\mathbf{j}}) \mathrm{ms}^{-1}\). The horizontal range of the projectile is (a) \(4.9 \mathrm{~m}\) (b) \(9.6 \mathrm{~m}\) (c) \(19.6 \mathrm{~m}\) (d) \(14 \mathrm{~m}\)

A projectile is projected with velocity \(k v_{e}\) vertically upward direction from the ground into the space \(\left(v_{e}\right.\) is the escape velocity and \(\left.k<1\right)\). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go will be \((R=\) radius of earth \() \quad\) [Orissa JEE 2010] (a) \(\frac{R}{k^{2}+1}\) (b) \(\frac{R}{k^{2}-1}\) (c) \(\frac{R}{1-k^{2}}\) (d) \(\frac{R}{k+1}\)

Two cars of masses \(m_{1}\) and \(m_{2}\) are moving in circles of radii \(r_{1}\) and \(r_{2}\) respectively. Their speeds are such that they make complete circles in the same time \(t\). The ratio of their centripetal acceleration is [AIEEE 2012] (a) \(m_{1} r_{1}: m_{2} 5\) (b) \(m_{1}: m_{2}\) (c) \(r_{1}: r_{2}\) (d) \(1: 1\)

A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of \(60 \mathrm{kmh}^{-1}\) and at a height of \(490 \mathrm{~m}\). At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target? (a) \(\frac{400}{3} \mathrm{~m}\) (b) \(\frac{500}{3} \mathrm{~m}\) (c) \(\frac{1700}{3} \mathrm{~m}\) (d) \(498 \mathrm{~m}\)

A body is projected with speed \(v \mathrm{~ms}^{-1}\) at angle \(\theta\). The kinetic energy at the highest point is half of the initial kinetic energy. The value of \(\theta\) is (a) \(30^{\circ}\) (b) \(45^{\circ}\) (c) \(60^{\circ}\) (d) \(90^{\circ}\)
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