Question

Ex 1: The speed with which a body is projected inside a tunnel is
40 ms-1 and the maximum range is 80√3 m. Find the height of
the tunnel.
(a) 5 m
(b) 10 m
(C) 15 m
(d) 20 m​

Answer 1

Answer:

answer is option D = 20 m

Answer 2

The Height of the tunnel, is (d) 20 m.

The following are the formulae that will be used:

1. To find the Angle $\theta$ in which the body gets projected inside, use this formulae -  $R = \frac{u^{2}\sin 2\Theta}{g}$
2. Finally, after we get the angle, we will utilise it to calculate the height using the following formulae - $h = \frac{u^{2}\sin ^{2}\Theta}{2g}$

Prerequisite Knowledge:

• The velocity with which the body is projected is called velocity of projection. It is denoted by ' u ' in the equation.
• The greatest distance an object can travel without taking dispersion into account is called maximum range of that object. It is denoted as ' R ' in the equation.
• Free-falling objects accelerate downward at a rate of $9.8 ms^{-2}$ (on Earth), this is known as Acceleration of Gravity and is denoted by ' g '.

Given in question:

• $u = 40 ms^{-1}$
• $R = 80\sqrt{3} m$
• $g = 9.8 ms^{-2}$ { On Earth } = $10 ms^{-2}$ [round off]

So,

$R = \frac{u^{2}\sin 2\Theta}{g}\\\Rightarrow 80\sqrt{3} = \frac {40 \times 40 \times \sin 2\Theta}{10}\\\Rightarrow \sin2\Theta = \frac {\sqrt{3}}{2}\\\Rightarrow 2\Theta = \sin^{-1}\frac {\sqrt{3}}{2}\\\Rightarrow \Theta = \frac{60\degree}{2}\\\Rightarrow \Theta = 30\degree$

Now we have the Angle of projection i.e. $\Theta = 30 \degree$, then:

$h = \frac{u^{2}\sin ^{2}\Theta}{2g}\\\Rightarrow h = \frac{(40)^{2}\times \sin ^{2}30\degree}{2\times10}\\\Rightarrow h = 20m$

Therefore,

The Height of the tunnel in which the body is projected is 20m.

Learn more about Maximum  Range of projections:

https://brainly.in/question/11523460?msp_srt_exp=4

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