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Chapter 2: Problem 91

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) \(60 \mathrm{~m}\)

Short Answer

Expert verified
The height at which the speed of the projectile becomes half of the initial speed is \(60 m\). Therefore, the correct answer is (D).

Step by step solution

01

Calculate the horizontal and vertical components of the initial velocity

In projectile motion, we need to find the horizontal and vertical components of the initial velocity. The horizontal component can be calculated using the formula: \(v_{x0} = v_0 \cos(\theta)\) And the vertical component can be calculated using the formula: \(v_{y0} = v_0 \sin(\theta)\) Here, v0 = \(40 ms^{-1}\), which is the initial velocity 胃 = \(60^\circ\), angle of projection w.r.t the horizontal We calculate the components: \(v_{x0} = 40 \times \cos(60^\circ) = 40 \times (1/2) = 20 ms^{-1}\) \(v_{y0} = 40 \times \sin(60^\circ) = 40 \times (\sqrt{3}/2) = 20\sqrt 3 ms^{-1}\)
02

Find the vertical component of final velocity

Since we are given that the final speed must be half of the initial speed, let's denote this final speed by \(v_{f}\). Therefore, \(v_{f} = 0.5v_{0}=20 ms^{-1}\) Since the projectile motion does not have any acceleration in the horizontal direction, \(v_{fx}=v_{x0}=20 ms^{-1}\) Now, we can use the Pythagorean relation: \(v_{f}^{2} = v_{fx}^{2} + v_{fy}^{2}\) Substitute the values: \(20^2 = 20^2 + v_{fy}^{2}\) Solving for the vertical component of the final velocity, we find that: \(v_{fy}^{2} = 0\) So, \(v_{fy} = 0 ms^{-1}\)
03

Calculate the height at which the speed becomes half

In this step, we will use the equations of motion in the vertical direction to find the height at which the speed of the particle becomes half. We will use the following equation: \(v_{fy}^2 = v_{y0}^2 - 2gH\) Here, \(v_{fy} = 0 ms^{-1}\) (from Step 2) \(v_{y0} = 20\sqrt 3 ms^{-1}\) (from Step 1) g = \(10 ms^{-2}\) (gravitational acceleration) H = the height we need to find Now, substitute the values and solve for H: \(0^2 = (20\sqrt 3)^2 - 2(10)H\) \(0 = 1200 - 20H\) We get, \(H = \dfrac{1200}{20} \) \(H = 60 m\) The height at which the speed of the projectile becomes half of the initial speed is 60 m. Therefore, the correct answer is (D).

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Most popular questions from this chapter

A projectile is given an initial velocity of \((\hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}\), where \(\hat{i}\) is along the ground and \(\hat{j}\) is along the vertical. If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), the equation of its trajectory is [2013] (A) \(y=2 x-5 x^{2}\) (B) \(4 y=2 x-5 x^{2}\) (C) \(4 y=2 x-25 x^{2}\) (D) \(y=x-5 x^{2}\)

A car is moving on a circular path of radius \(100 \mathrm{~m}\). Its speed \(v\) is changing with time as \(v=2 t^{2}\), where \(v\) in \(\mathrm{ms}^{-1}\) and \(t\) in second. The acceleration of car at \(t=5 \mathrm{~s}\) is approximately (A) \(20 \mathrm{~ms}^{-1}\) (B) \(25 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(32 \mathrm{~ms}^{-1}\)

A particle is projected with a velocity \(v\) such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where \(g\) is acceleration due to gravity) (A) \(\frac{4 v^{2}}{5 g}\) (B) \(\frac{4 g}{5 v^{2}}\) (C) \(\frac{v^{2}}{g}\) (D) \(\frac{4 v^{2}}{\sqrt{5} g}\)

The displacement of a particle is given by \(x=(t-2)^{2}\), where \(x\) is in metres and \(t\) in seconds. The distance covered by the particle in first \(4 \mathrm{~s}\) is (A) \(4 \mathrm{~m}\) (B) \(8 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(16 \mathrm{~m}\)

A stone is projected from the ground with velocity \(50 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ} .\) It crosses a wall after \(3 \mathrm{~s}\). How far beyond the wall the stone will strike the ground \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(90.2 \mathrm{~m}\) (B) \(89.6 \mathrm{~m}\) (C) \(86.6 \mathrm{~m}\) (D) \(70.2 \mathrm{~m}\)
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