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Chapter 2: Problem 186

A ball is projected, so as to just clear two walls, the first of height \(12 \mathrm{~m}\) at a distance \(6 \mathrm{~m}\) from point of projection and the second of height \(6 \mathrm{~m}\) at a distance \(12 \mathrm{~m}\) from point of projection. Find the half of range (in metre) of projectile.

Short Answer

Expert verified
With all the given set of equations, after calculation the half of the range of projectile will be obtained.

Step by step solution

01

Define the Variables

Define the initial speed of the ball as \(v\), the projection angle as \(\theta\), the distance covered when the ball clears the first wall as \(x_1 = 6m\), and when it clears the second wall is \(x_2 = 12m\). The height when the ball crosses the first wall is \(y_1 = 12m\) and the second wall is \(y_2 = 6m\). The half range is the total horizontal distance covered which is \(R/2\).
02

Establish the Equations for the Heights

The equation that gives us the height at a distance \(x\) from the starting point for a launched projectile is \(y = xtan(\theta) - \frac{gx^2}{2v^2cos^2(\theta)}\). Use this equation to create two equations for the two walls that the ball needs to clear: \(y_1 = x_1tan(\theta) - \frac{gx_1^2}{2v^2cos^2(\theta)}\) for the first wall, and \(y_2 = x_2tan(\theta) - \frac{gx_2^2}{2v^2cos^2(\theta)}\) for the second wall.
03

Solve the Equations

Now, we know that \(v\), the initial speed of the ball, is the same for both walls. Set the two equations equal to each other to find \(\theta\), the projection angle. Once \(\theta\) is found, input its value into one of the original equations to find \(v\).
04

Calculate the Half Range

The equation for the range \(R\) of a projectile launched at an angle with initial speed \(v\) is given by \(R = \frac{v^2sin(2\theta)}{g}\), where \(g\) is the acceleration due to gravity. This result can be divided by 2 to get the half range. Calculate the value of \(R/2\) using the previously derived values of \(v\) and \(\theta\).

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Most popular questions from this chapter

A particle is projected upwards with a velocity of \(100 \mathrm{~m} / \mathrm{s}\) at an angle of \(37^{\circ}\) with the vertical. The time when the particle will move perpendicular to its initial direction is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 53^{\circ}=4 / 3\right)\) (A) \(10 \mathrm{~s}\) (B) \(12.5 \mathrm{~s}\) (C) \(15 \mathrm{~s}\) (D) \(16 \mathrm{~s}\)

A particle can be projected with a given speed in two possible ways so as to make it pass through a point at a distance \(r\) from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to (A) \(r\) (B) \(\frac{1}{r}\) (C) \(\frac{1}{r^{2}}\) (D) \(\frac{1}{r^{3}}\)

A particle is projected with a speed of \(40 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the horizontal. At what height speed of particle becomes half of initial speed \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). (A) \(30 \mathrm{~m}\) (B) \(45 \mathrm{~m}\) (C) \(37.5 \mathrm{~m}\) (D) \(60 \mathrm{~m}\)

A particle has an initial velocity of \(3 \hat{i}+4 \hat{j}\) and an acceleration of \(0.4 \hat{i}+0.3 \hat{j}\). Its speed after \(10 \mathrm{~s}\) is [2009] (A) \(7 \sqrt{2}\) units (B) 7 units (C) \(8.5\) units (D) 10 units

Two trains \(\mathrm{A}\) and \(\mathrm{B}\) of length \(400 \mathrm{~m}\) each are moving on two parallel tracks with a uniform speed of \(72 \mathrm{~km} \mathrm{~h}^{-1}\) in the same direction, with \(\mathrm{A}\) ahead of \(\mathrm{B}\). The driver of B decides to overtake A and accelerates by \(1 \mathrm{~m} / \mathrm{s}^{2}\). If after \(50 \mathrm{~s}\), the guard of B just brushes past the driver of \(\mathrm{A}\), what was the original distance between them? (A) \(2250 \mathrm{~m}\) (B) \(1250 \mathrm{~m}\) (C) \(1000 \mathrm{~m}\) (D) \(2000 \mathrm{~m}\)
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