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The velocity of projection of oblique projectile is $(6 \hat{i}+8 \widehat{j}) m s^{-1} .$ The horizontal range of the projectile is
A.$4.9 \mathrm{~m}$
C.$19.6 \mathrm{~m}$
D.$14 \mathrm{~m}$

Last updated date: 20th Jun 2024
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Hint: A projectile is an object that we give, and gravity acts on an initial velocity. The horizontal range is the distance along the horizontal plane of a projectile. In addition, it would travel before it reached the same vertical position from which it began. Split the vector into its X an Y coordinates and Calculate R using Newton's law of motion.

Formula used:
$\text{R}=\dfrac{{{\text{v}}^{2}}\sin 2\theta }{\text{g}}$

Complete Step-by-Step solution:
The projectile's horizontal motion is the result of any object in motion's tendency to remain in motion at constant velocity. Because of the absence of horizontal forces, a projectile with a constant horizontal velocity stays in motion.

After that, the horizontal range is depending upon the initial velocity V0, the launch angle θ, and the acceleration occurring due to the gravity.
$\overrightarrow{\mathrm{v}}=6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}$

Comparing with $\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$ we get
$\mathrm{v}_{\mathrm{x}}=6 \mathrm{~ms}^{-1}$ and $\mathrm{v}_{\mathrm{y}}=8 \mathrm{~ms}^{-1}$
also $\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}=36+64=100$
or $\mathrm{v}=10 \mathrm{~ms}^{-1}$
$\sin \theta=\dfrac{8}{10}$ and $\cos \theta=\dfrac{6}{10}$
$\mathrm{R}=\dfrac{\mathrm{v}^{2} \sin 2 \theta}{\mathrm{g}}=\dfrac{2 \mathrm{v}^{2} \sin \theta \cos \theta}{\mathrm{g}}$
$\text{R}=2\times 10\times 10\times \dfrac{8}{10}\times \dfrac{6}{10}\times \dfrac{1}{10}$
$\therefore \text{R}=9.6~\text{m}$
The horizontal range of the projectile is: $9.6~\text{m}$

Hence, the correct option is (B).

The greater the release level, the greater the distance covered in flight. This is because the greater the release of the projectile, the longer it will be in the air. The horizontal component will act for longer on the projectile. In order to provide the equation for the horizontal distance, the slope percent equation can be rearranged. Rearrange equation terms: multiply both sides by running. Split both sides by slope percentage. The initial launch angle will be anywhere from 0 to 90 degrees for an object launched into projectile motion.