Question

Horizontal range of projectile

Answer 1

Answer:

• The horizontal range (R) of the projectile is 30 m.

Given:

1. The Equation of projectile is y = x √3  - x² / 10 √3

Explanation:

$\rule{300}{1.5}$

As the question states that we need to find the Horizontal range of the projectile.

Now, From the given equation,

⇒ y = x √3 - x² / 10 √3

For getting the maximum range the vertical distance should be zero i.e y = 0 m and x = R (Maximum range)

Here, we took y as zero because when body completes it maximum horizontal range, the net displacement in y - direction is zero.

Now, Substituting,

⇒ 0 = x √3 - x² / 10 √3

⇒ x √3 = x² / 10 √3

Cancelling the common term

⇒ √3 = x / 10 √3

Rearranging,

⇒ √3 × 10 √3 = x

⇒ x = 10 √3 × √3

⇒ x = 10 × 3

⇒ x = 30

∵ [ x = R ]

⇒ R = 30 m.

∴ The horizontal range (R) of the projectile is 30 m.

$\rule{300}{1.5}$

Answer 2

$</p><p>\tt{\huge {\pink{Hello!!! }}}$

$</p><p>\tt{\red{Given \: Equation \: Of \: Trajectory \: Is - }}$

$\tt{ \purple{y \: = > \: \sqrt{3} x - \frac{ {x}^{2} }{10 \sqrt{3} } }}$

$</p><p>\tt{\orange{To \: Find \: Horizontal \: Range - }}$

For horizonal range R, the value of Y is zero . Hence in the equation is trajectory , we need to substitute the value of Y as Zero.

The Value Of X Obtained becomes the Horizontal Range Of The Given Projectile.

Hence,

$</p><p>\tt{\blue{0 \: = > \: \sqrt{3} x - \frac{ {x}^{2} }{10 \sqrt{3} } }}$

$</p><p>\tt{\red{= > \: \sqrt{3} x =\frac{ {x}^{2} }{10 \sqrt{3} } }}$

$</p><p>\tt{\orange {=> X = 10\sqrt{3} \times \sqrt{3} = 30 }}$

$</p><p>\tt{\huge{\boxed {\boxed {\purple{\therefore {R = 30 m. }}}}}}$