/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 133 The horizontal range of a projec... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 133

The horizontal range of a projectile fired at an angle of 15 is \(50 \mathrm{~m}\). If it is fired with the same speed at an angle of 45 , its range will be (A) \(60 \mathrm{~m}\) (B) \(71 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(141 \mathrm{~m}\)

Short Answer

Expert verified
The horizontal range of a projectile fired at an angle of 45 degrees with the same initial speed is approximately \(100 \mathrm{~m}\). Therefore, the correct answer is (C) 100 meters.

Step by step solution

01

Understand projectile motion and range formula

Projectile motion involves two independent motions: horizontal and vertical motion. They can be analyzed separately. The horizontal range (R) of a projectile is given by the formula: \[ R = \frac{v_i^2 \sin(2\theta)}{g} \] Where \(v_i\) is the initial speed, \(\theta\) is the angle of the projectile, and \(g\) is the acceleration due to gravity (approximately 9.81 m/s虏).
02

Set up the equation for the 15-degree angle

Given that the range at a 15-degree angle is 50 meters, we can write the equation as: \[ 50 = \frac{v_i^2 \sin(2(15))}{9.81} \] We need to find the initial speed, \(v_i\), to calculate the range at a 45-degree angle.
03

Solve for the initial speed, \(v_i\)

Rearrange the equation to solve for the initial speed, \(v_i\): \[ v_i = \sqrt{\frac{50 \times 9.81}{\sin(2(15))}} \] Calculating the initial speed, we get: \[ v_i \approx 42.61 \mathrm{~m/s} \]
04

Set up the equation for the 45-degree angle

Now that we have the initial speed, we can calculate the range at a 45-degree angle using the range formula: \[ R = \frac{(42.61)^2 \sin(2(45))}{9.81} \]
05

Calculate the range

Calculate the range for a 45-degree angle: \[ R \approx 100.19 \mathrm{~m} \] Since this is close to 100 meters, the correct answer is (C) 100 meters.

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Most popular questions from this chapter

A boy playing on the roof of a \(10 \mathrm{~m}\) high building throws a ball at a speed of \(10 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal. How far from the throwing point will the ball be at the height of \(10 \mathrm{~m}\) from the ground? \(\left[g=10 \mathrm{~m} / \mathrm{s}^{2}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\) \([\mathbf{2 0 0 3}]\) (A) \(5.20 \mathrm{~m}\) (B) \(4.33 \mathrm{~m}\) (C) \(2.60 \mathrm{~m}\) (D) \(8.66 \mathrm{~m}\)

A projectile is thrown horizontally from top of a building of height \(10 \mathrm{~m}\) with certain speed \((u)\). At the same time another projectile is thrown from ground \(10 \mathrm{~m}\) away from the building with equal speed \((u)\) on the same vertical plane. If they collide after \(2 s\), then choose the correct options. (A) The angle of projection for second projectile is \(60^{\circ}\) and \(u=10 \mathrm{~ms}^{-1}\) (B) The angle of projection for second projectile is \(90^{\circ}\) and \(u=5 \mathrm{~ms}^{-1}\) (C) The angle of projection for second projectile is \(60^{\circ}\) and \(u=5 \mathrm{~ms}^{-1}\) (D) The angle of projection for second projectile is \(45^{\circ}\) and \(u=10 \mathrm{~ms}^{-1}\)

A shell fired from the ground is just able to cross in a horizontal direction the top of a wall \(90 \mathrm{~m}\) away and \(45 \mathrm{~m}\) high. The direction of projection of the shell is (A) \(25^{\circ}\) (B) \(30^{\circ}\) (C) \(60^{\circ}\) (D) \(45^{\circ}\)

A stone is thrown with speed of \(20 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the ground. Speed of stone when makes an angle of \(30^{\circ}\) with the horizontal is (A) \(10 \mathrm{~m} / \mathrm{s}\) (B) \(10 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (C) \(\frac{20}{\sqrt{3}} \mathrm{~m} / \mathrm{s}\) (D) None of these

A stone tied to the end of a string \(80 \mathrm{~cm}\) long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in \(25 \mathrm{~s}\), what is the magnitude of acceleration of the stone? (A) \(9.91 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(19.82 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(31 \mathrm{~m} / \mathrm{s}^{2}\) (D) None
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