Question 146: Motion in 2D & Vectors

Question

Two bodies are projected from ground with same speeds $40 \mathrm{~ms}^{-1}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^{\circ}$, with horizontal then sum of the maximum heights, attained by the two projectiles, is $\mathrm{m}$. (Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

Accepted Answer

When two bodies are projected from the ground at the same speed of $40 \mathrm{~ms}^{-1}$ but at different angles, and they achieve the same range, we can derive the following:
Given that one projectile is launched at an angle of $60^\circ$ with respect to the horizontal, let's denote the angles of projection as $	heta_1$ and $	heta_2$. For the ranges to be equal, we know:
$ 	heta_1 + 	heta_2 = 90^\circ $
Since $	heta_1 = 60^\circ$, we can find $	heta_2$ as:
$ 	heta_2 = 30^\circ $
Next, to find the sum of the maximum heights attained by both projectiles, we use the formula for the maximum height $H_{\max}$:
$ H_{\max} = \frac{u^2 \sin^2 	heta}{2g} $
For the first body projected at $60^\circ$:
$ \left(H_{\max}ight)_1 = \frac{(40)^2 \sin^2 60^\circ}{2 	imes 10} $
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$ \left(H_{\max}ight)_1 = \frac{1600 	imes \left(\frac{\sqrt{3}}{2}ight)^2}{20} = \frac{1600 	imes \frac{3}{4}}{20} = \frac{1200}{20} = 60 \mathrm{~m} $
For the second body projected at $30^\circ$:
$ \left(H_{\max}ight)_2 = \frac{(40)^2 \sin^2 30^\circ}{2 	imes 10} $
Since $\sin 30^\circ = \frac{1}{2}$:
$ \left(H_{\max}ight)_2 = \frac{1600 	imes \left(\frac{1}{2}ight)^2}{20} = \frac{1600 	imes \frac{1}{4}}{20} = \frac{400}{20} = 20 \mathrm{~m} $
Therefore, the sum of the maximum heights attained by both projectiles is:
$ \left(H_{\max}ight)_1 + \left(H_{\max}ight)_2 = 60 \mathrm{~m} + 20 \mathrm{~m} = 80 \mathrm{~m} $

Motion in 2D & Vectors

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