First slide

Projection Under uniform Acceleration

Question

The velocity of projectile at the point of projection is $2 \hat{i} + 3 \hat{j}$ m/s. The velocity at a point where the projectile reaches the horizontal line through the point of projection is

Easy

A

$- 2\hat i + 3\hat j\,\,\,m/s$
B

$2\hat i - 3\hat j\,\,\,m/s$
C

$2\hat i + 3\hat j\,\,\,m/s$
D

$- 2\hat i - 3\hat j\,\,\,m/s$

Solution

Horizontal component of velocity remains unchanged but at the point where the projectile crosses the horizontal line through point of projection, the vertical component of velocity reverses its direction, so the velocity becomes $2\hat i - 3\hat j\,\,\,m/s$

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