Question

A ball is acted upon by the following velocities (i) 3 metre per sec due east (ii) 11 metre per sec due to south and (iii) 5 root 2 metre per sec due north-east. Find the magnitude and direction of resultant velocity.

Answer 1

This question can be easily solved with help of vector method .

let's see how ,

you should keep in mind ,

positive x-axis (east) = i

negative x-axis (west) = -i

positive Y-axis (north) = j

negative y-axis (south) = j

and north - east direction means, vector 45° with x-axis . so, (cos45° i + sin45° j)

now let's apply all above concept here.

A ball is acted upon by the following velocities

• (i) 3 metre per sec due east i.e., $v_1$ = 3i
• (ii) 11 metre per sec due to south i.e., $v_2$ = -11j
• (iii) 5 root 2 metre per sec due north-east. i.e., $v_3$= 5√2(cos45° i + sin45° j) = (5i + 5j)

now resultant velocity of ball = $v_1+v_2+v_3$

= 3i + (-11j) + (5i + 5j)

= 8i - 6j

so, magnitude of resultant velocity of ball = √{8² + (-6)²} = 10m/s

direction of resultant velocity, tanα = |-6/8| = tan37° ⇒ α = 37°