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Question: The velocity of a projectile at the initial point A is (2i + 3j) m/s. It's velocity (in m/s) at poin...

The velocity of a projectile at the initial point A is (2i + 3j) m/s. It's velocity (in m/s) at point B is

(A) -2i - 3j
(B) -2i + 3j
(C) 2i -3j
(D) 2i+3j

Explanation

Solution

This is the case of projectile motion. The velocity has 2 components, one along the x axis and the other along the y-axis. The body does not undergo any resistance in the x-direction. In the y direction, it moves against gravity. As gravity is a conservative force, the magnitude of the y component of velocity will stay the same.

Complete- step-by-step solution
Let’s split the motion into both the x and y components.
Y-component: When the body is going up, it moves opposite to the direction of gravity and experiences acceleration of ‘–g’ and is the vertical component of the velocity reduced to zero, this position is known as max height. When returning or losing altitude, it gains velocity (with acceleration ‘+g’) as it is moving in the direction of gravity. As gravity is a conservative force, hence after traveling equal distance up and falling down the same distance the y-component of the velocity will remain the same in magnitude but will reverse the sign (or reverse the direction). So the component will be -3 m/s
X-component: There is no force/acceleration experienced in this direction, hence the component is unchanged. Hence the component will be 2 m/s

Therefore the correct answer is option C

Note: If the velocity is given in terms of variables and we need to find the length of the path, then we will have to differentiate the velocity. Also, point B in this question is called the range of the projectile motion. It is the farthest distance that the body travels.