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Chapter 2: Problem 71

A particle is projected with a velocity \(u\), at an angle \(\alpha\), with the horizontal. Time at which its vertical component of velocity becomes half of its net speed at the highest point will be (A) \(\frac{u}{2 g}\) (B) \(\frac{u}{2 g}(\sin \alpha-\cos \alpha)\) (C) \(\frac{u}{2 g}(2 \cos \alpha-\sin \alpha)\) (D) \(\frac{u}{2 g}(2 \sin \alpha-\cos \alpha)\)

Short Answer

Expert verified
The time at which the vertical component of the velocity becomes half of its net speed at the highest point is given by: \(t=\frac{u(\sin\alpha-\cos\alpha)}{2g}\) The correct answer is option (B).

Step by step solution

01

Identify Initial Components of Velocity

First, we need to identify the initial components of velocity. We know that the angle of projection is given as \(\alpha\) and the initial velocity is given as \(u\). To find the initial vertical component of velocity, we can use the following formula: \(v_{0y} = u \sin \alpha\) Similarly, we can find the initial horizontal component of the velocity: \(v_{0x} = u \cos \alpha\)
02

Calculate Net Speed at Highest Point

In order to find the net speed at the highest point, we need to find the vertical component of velocity and horizontal component of velocity at that point. At the highest point, the vertical component of velocity becomes 0, so the net speed is equal to the horizontal component of velocity. So: \(v_{net} = v_{0x} = u \cos \alpha\)
03

Calculate Half of Net Speed at Highest Point

Now, calculate half of the net speed at the highest point: \(\frac{1}{2}v_{net} = \frac{1}{2}(u \cos \alpha) = \frac{u \cos \alpha}{2}\)
04

Formula for Vertical Component of Velocity at Time t

Next, we need to determine the formula for the vertical component of velocity at any given time. The formula is: \(v_y(t) = v_{0y} - gt\) where \(g\) is the gravitational acceleration.
05

Solve for the Time t

Now set the formula for the vertical component of the velocity at time t equal to half the net speed at the highest point and solve for the time, t: \(\frac{u \cos \alpha}{2} = u \sin \alpha - gt\) Rearrange the equation to isolate t: \(t=\frac{u(\sin\alpha-\cos\alpha)}{2g}\) So, the time at which the vertical component of the velocity becomes half of its net speed at the highest point is given by: \(t=\frac{u(\sin\alpha-\cos\alpha)}{2g}\) The correct answer is option (B).

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