Question

A projectile is thrown into space so as to have the maximum possible horizontal range equal to
400 m. Taking the point of projection as the origin, the coordinates of the point where the velocity
of the projectile is minimum, are
(A) (400, 100)
(B) (200, 100)
(C) (400, 200)
(D) (200, 200)

Answer the question with steps and u will be marked as the BRAINLIEST!! ​

Answer 1

Answer:

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\red{\mathfrak{Given}}}}}}}}}$

Maximum Horizontal Range = 400 m

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\red{\mathfrak{To \:Find}}}}}}}}}$

Coordinate of the point where the projectile had "Least Velocity"

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\red{\mathfrak{Concept}}}}}}}}}$

Any projectile has Least Velocity at its highest point. This is because , at the highest point ,the projectile has only horizontal component of velocity

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\red{\mathfrak{Derivation}}}}}}}}}$

Range = R = u²sin(2θ)/g

For maximum R , 2θ = 90°

=> θ = 45°,

then R (max) = (u²/g)

Now, maximum height

H = u² sin²(θ)/2g

=> H = u² sin²(45°)/2g

=> H = u²/4g

=> H = R/4.

So the maximum height is 400/4 = 100 m

Again maximum height is obtained at half of the range (i.e at 200 m)

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\red{\mathfrak{Final \:Answer}}}}}}}}}$

The coordinate (x,y) is (200,100).

Answer 2

$\bf{\huge{\underline{\boxed{\sf\purple{Answer:200\:, 100}}}}}$

$\rule{100}2$

$\textbf{\underline{\underline{Explanation:}}}$

As we know, Velocity remains minimum at the highest point.

It is given that,

→ The maximum possible horizontal range = 400 m.

Now,

It's x co-ordinate :-

$= \dfrac{400}{2}$

= 200 _______________ ( ! )

➡For y co - ordinate (maximum height )

Maximum height

$= \dfrac{u^2\sin^2 \times 45}{2g}$

$=\dfrac{u^2}{4g}$

$= \dfrac{1}{4} \times 400$

= 100 ____________ ( !! )

∴ It's co - ordinates are - ( 200, 100 )

$\rule{200}2$