Physics · 2019
JEE Main · 9 January 2019, Shift 1 · Q2
A particle is moving with a velocity v = K (y i +x j ), where K is a constant. The general equation for its path is:
A particle is moving with a velocity
$$\vec{v}=\mathrm{K}(y \hat{i}+x \hat{j}) \text {, where } \mathrm{K} \text { is a constant. }
$$
The general equation for its path is :
Official answer
(1)
$\displaystyle y^2=x^2+\text { constant }$
More from Motion in Two and Three Dimensions
- The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is:2023
- A stone is projected at angle 30° to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the…2023
- A ball of mass 100 g is projected with velocity 20 m / s at 60^° with horizontal. The decrease in kinetic energy of the ball during the motion from…2025
- A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal, then ' tan θ…2022
- The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ____ m.2024
- A projectile is thrown with initial velocity of (3 i+4 j) m / s. Here ' i ' is along the horizontal direction and ' j ' is assumed in vertical…2021
- The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is (√3)/2 u. The time of…2023
- A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of 100 m from the foot of…2024
JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.