Kinematics Question 628
Question: A projectile of mass m is thrown with a velocity v making an angle $ 60{}^\circ $ with the horizontal. Neglecting air resistance, the change in velocity from the departure A to it’s arrival at B, along the vertical direction is
Options:
A) 2v
B)$ \sqrt{3}\text{v} $
C) v
D)$ \frac{\text{v}}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] At points A and B the vertical component of velocity is $ v\sin 60{}^\circ $ but their directions are opposite.
Hence, change in velocity. $ \Delta v=v\sin 60{}^\circ -( -v\sin 60{}^\circ)=2v\sin 60{}^\circ =\sqrt{3}v $
BETA
