Question

the magnitude and direction of resultant velocities of 100m/s and 60 m/s inclined to each other at angle 60°​

Answer 1

Answer:

Given A=100m/s

B=60m/s

angle between A and B=60 degrees

$resultant = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos(x)} \\ = \sqrt{ {100}^{2} + {60}^{2} + 2(100)(60) \cos( {60}^{0} ) } \\ = \sqrt{13600 + 6000} \\ = \sqrt{19600} \\ = 140m {s}^{ - 2} \\ \\ x = { \tan }^{ - 1} \frac{b \: \sin(x) }{a + b \cos(x) } \\ = { \tan }^{ - 1} \frac{60 \frac{ (\sqrt{3} }{2}) }{100 + 60( \frac{1}{2} )} \\ = { \tan }^{ - 1} \frac{3 \sqrt{3} }{13} with \: a$