Question

A projectile is thrown into space to have the maximum possible horizontal range equal to 400 m. Taking the point of a projectile as the origin, the coordinates of the point where the velocity of the projectile is minimum
1-(400,100)
2-(200,100)
3-(400,200)
4-(200,200)

Answer 1

Hint: To have the maximum range the angle with the horizontal is always \[45{}^\circ \]. In projectile problems when the body is projected whether, from the ground or a certain height, there is no acceleration in the horizontal direction. There is constant acceleration in the vertical direction which is always directed downwards and this is constant. This is the acceleration due to gravity.

Complete step by step answer:
Now the projectile is thrown into space and we are given that it achieves its maximum possible range which is 400m. Taking the launching of the projectile from the origin, let us assume that the initial velocity of the projectile is u.
To achieve the maximum range the launching angle of the projectile is always \[45{}^\circ \]
\[R=\dfrac{{{u}^{2}}\sin 2\theta }{g}\]
Substituting the values,
\[400=\dfrac{{{u}^{2}}\sin 90}{9.8}\]
We get u=62.6 m/s
Now we know that velocity of the projectile is minimum at maximum height,
\[H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
Substituting the values
 \[H=\dfrac{{{60.6}^{2}}{{\sin }^{2}}45}{2\times 9.8}\]
H=100m
Also, maximum height takes place at the middle point of the maximum range that is 200m
So, the coordinates are (200,100)

So, the correct answer is “Option 2”.

Note:
We can arrive at the solution directly also. Since we know that the Speed of the particle will be minimum at the highest point of the parabola. So, the coordinate of the highest point will be \[(\dfrac{R}{2},\dfrac{R}{4})\].