Answer
396.9k+ views
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).
Note:
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.
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).
Note:
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)