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Chapter 5: Problem 75

Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is \(\pi / 3\) and its maximum height is \(h_{1}\) then the maximum height of the other will be a. \(3 h_{1}\) b. \(2 h_{1}\) c. \(h_{1} / 2\) d. \(h_{1} / 3\)

Short Answer

Expert verified
The maximum height of the other stone is \( h_1/3 \).

Step by step solution

01

Understanding Range and Angle Relationship

When two projectiles have the same initial speed and cover the same range, the angles of projection, \( \theta_1 \) and \( \theta_2 \), will add up to \( \pi/2 \). Here, one of the angles is \( \pi/3 \). Hence, the other angle, \( \theta_2 \), can be calculated as follows: \[ \theta_2 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}. \]
02

Maximum Height Formula

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

Calculating Maximum Height \( h_1 \)

For the first stone with \( \theta_1 = \pi/3 \), the maximum height \( h_1 \) is \[ h_1 = \frac{v^2 \sin^2(\pi/3)}{2g} = \frac{v^2 \times \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3v^2}{8g}. \]
04

Calculating Maximum Height \( h_2 \)

For the second stone with \( \theta_2 = \pi/6 \), the maximum height \( h_2 \) is \[ h_2 = \frac{v^2 \sin^2(\pi/6)}{2g} = \frac{v^2 \times \left(\frac{1}{2}\right)^2}{2g} = \frac{v^2}{8g}. \]
05

Comparing \( h_1 \) and \( h_2 \)

The relation between \( h_1 \) and \( h_2 \) can be calculated as follows: \[ h_2 = \frac{v^2}{8g} = \frac{1}{3} \times \frac{3v^2}{8g} = \frac{1}{3}h_1. \] Therefore, \( h_2 = \frac{h_1}{3}. \)

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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 the horizontal distance it covers during its motion. This is a crucial aspect to understand when studying projectile motion. A projectile's range depends on several key factors:
  • Initial speed: The greater the speed, the further the projectile can travel.
  • Angle of projection: The range is maximum when the angle is 45 degrees if air resistance is neglected.
  • Gravitational acceleration: On Earth, this constant typically measured as 9.81 m/s虏 always acts downwards.
The interesting property of the range is its dependency on the angle of projection. For a given initial speed and ignoring air resistance, two angles of projection will result in the same range if they add up to 90 degrees or \(\pi/2\) radians. Therefore, if one of the angles is \(\pi/3\), the other must be \(\pi/6\) to satisfy this condition. This understanding helps us solve problems where different projectiles have the same range but different angles.
Angle of Projection
Angle of projection is the angle at which a projectile is launched with respect to the horizontal axis. It plays a critical role in determining how far and how high a projectile will travel. In relation to the range, we know:
  • If two projectiles have the same initial speed and range, their angles of projection must add up to \(\pi/2\) radians.
  • The angle affects both the height and the range of the projectile, crucially determining the trajectory.
When solving problems of this nature, use the relationship that if one angle is \(\pi/3\), the complementary angle will be \(\pi/6\). Each angle affects not just the maximum height the projectile can reach but also influences the range. Thus, understanding the angle of projection can help in predicting the projectile鈥檚 behavior accurately.
Maximum Height
Maximum height is the highest vertical point reached by a projectile in its trajectory and can be calculated using the formula:\[ h = \frac{v^2 \sin^2 \theta}{2g} \]where:
  • \(v\) is the initial velocity
  • \(\theta\) is the angle of projection
  • \(g\) is the gravitational acceleration (9.81 m/s虏)
In this exercise, we explore two projectiles with different angles but the same initial speed giving us a particular physics insight. For one projectile at \(\pi/3\), the maximum height \(h_1\) is calculated as \(\frac{3v^2}{8g}\). For the other at \(\pi/6\), \(h_2\) is \(\frac{v^2}{8g}\).
By comparison, \(h_2 = \frac{h_1}{3}\), showing how significant the angle of projection is when determining maximum height. Therefore, the angle can greatly affect the outcome of the height achieved during projectile motion.

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Most popular questions from this chapter

i. The maximum height attained by a projectile is increased by \(5 \%\). Keeping the angle of projection constant, what is the percentage increase in horizontal range? a. \(5 \%\) b. \(10 \%\) c. \(15 \%\) d. \(20 \%\) The maximum height attained by a projectile is increased by \(10 \%\). Keeping the angle of projection constant, what is the percentage increase in the time of flight? a. \(5 \%\) b. \(10 \%\) c. \(20 \%\) d. \(40 \%\)

Ship \(\mathrm{A}\) is travelling with a velocity of \(5 \mathrm{kmh}^{-1}\) due east. The second ship is heading \(30^{\circ}\) east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship? a. \(10 \mathrm{kmh}^{-1}\) b. \(9 \mathrm{kmh}^{-1}\) c. \(8 \mathrm{kmh}^{-1}\) d. \(7 \mathrm{kmh}^{-1}\)

A number of bullets are fired in all possible directions with the same initial velocity \(u\). The maximum area of ground covered by bullets is a. \(\pi\left(\frac{u^{2}}{g}\right)^{2}\) b. \(\pi\left(\frac{u^{2}}{2 g}\right)^{2}\) c. \(\pi\left(\frac{u}{g}\right)^{2}\) d. \(\pi\left(\frac{u}{2 g}\right)^{2}\)

A car is moving towards cast with a speed of \(25 \mathrm{~km} / \mathrm{h}\). To the driver of the car, a bus appears to move towards north with a speed of \(25 \sqrt{3} \mathrm{~km} / \mathrm{h}\). What is the actual velocity of the bus? a. \(50 \mathrm{~km} / \mathrm{h}, 30^{\circ}\) cast of north b. \(50 \mathrm{~km} / \mathrm{h}, 30^{\circ}\) north of east c. \(25 \mathrm{~km} / \mathrm{h}, 30^{\circ}\) east of north d. \(25 \mathrm{~km} / \mathrm{h}, 30^{\circ}\) north of east

A bird is fiying towards north with a velocity \(40 \mathrm{~km} / \mathrm{h}\) and a train is moving with a velocity \(40 \mathrm{~km} / \mathrm{h}\) towards east. What is the velocity of the bird noted by a man in the train? a. \(40 \sqrt{2} \mathrm{~km} / \mathrm{h} \mathrm{N}-\mathrm{E}\) b. \(40 \sqrt{2} \mathrm{~km} / \mathrm{h} \mathrm{S}-\mathrm{E}\) c. \(40 \sqrt{2} \mathrm{~km} / \mathrm{h} \mathrm{N}-\mathrm{W}\) d. \(40 \sqrt{2} \mathrm{~km} / \mathrm{h} \mathrm{S}-\mathrm{W}\)
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