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Chapter 2: Problem 89

The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is (A) \(\theta\) (B) \(\tan ^{-1}(\theta / t)\) (C) \(\tan ^{-1}\left(\frac{v \cos \theta}{v \sin \theta-g t}\right)\) (D) \(\tan ^{-1}\left(\frac{v \sin \theta-g t}{v \cos \theta}\right)\)

Short Answer

Expert verified
The short answer is: \(\alpha = \tan^{-1}\left(\frac{v\sin(\theta) - gt}{v\cos(\theta)}\right)\). The correct option is (D).

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

The horizontal component of the initial velocity is given by: \(v_x = v\cos(\theta)\) The vertical component of the initial velocity is given by: \(v_y = v\sin(\theta)\)
02

Find the horizontal and vertical components of the velocity at time t

The horizontal component of the velocity remains constant throughout the motion, so: \(v_{xt} = v\cos(\theta)\) The vertical component of the velocity at time t can be found using the equation: \(v_{yt} = v_y - gt\) where g is the acceleration due to gravity. Substituting the expression for \(v_y\), we get: \(v_{yt} = v\sin(\theta) - gt\)
03

Find the angle that the velocity vector makes with the horizontal at time t

We will use the inverse tangent function to find the angle. Let 伪 be the angle the velocity vector makes with the horizontal at time t. We have: \(\tan(\alpha) = \frac{v_{yt}}{v_{xt}}\) Substituting the expressions for \(v_{xt}\) and \(v_{yt}\), we get: \(\tan(\alpha) = \frac{v\sin(\theta) - gt}{v\cos(\theta)}\) Now, we will find the angle 伪 using the inverse tangent function: \(\alpha = \tan^{-1}\left(\frac{v\sin(\theta) - gt}{v\cos(\theta)}\right)\) Comparing with the given options, we can see that the correct answer is: (D) \(\tan^{-1}\left(\frac{v\sin(\theta) - gt}{v\cos(\theta)}\right)\)

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Most popular questions from this chapter

From the top of a wall of height \(5 \mathrm{~m}\), a ball is thrown horizontally with speed of \(6 \mathrm{~m} / \mathrm{s}\). How far from the wall will the ball land?

The maximum height attained by a projectile is increased by \(5 \%\), keeping the angle of projection constant. The corresponding percentage increase in horizontal range will be (A) \(5 \%\) (B) \(10 \%\) (C) \(15 \%\) (D) \(20 \%\)

A road is \(5 \mathrm{~m}\) wide. Its radius of curvature is \(20 \sqrt{6} \mathrm{~m}\). The outer edge is above the inner edge by a distance of \(1 \mathrm{~m}\). This road is most suited for a speed \(\left(g=10 \mathrm{~ms}^{-2}\right.\) ) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (C) \(100 \mathrm{~ms}^{-1}\) (D) \(40 \sqrt{6} \mathrm{~ms}^{-1}\)

Two bodies are projected at angles \(\theta\) and \((90-\theta)\) to the horizontal with the same speed. The ratio of their times of flight is (A) \(\sin \theta: 1\) (B) \(\cos \theta: 1\) (C) \(\sin \theta: \cos \theta\) (D) \(\cos \theta: \sin \theta\)

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