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 its arrival at $$B,$$ along the vertical direction is
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 its arrival at $$B,$$ along the vertical direction is
A.
$$2v$$
B.
$$\sqrt 3 v$$
C.
$$v$$
D.
$$\frac{v}{{\sqrt 3 }}$$
Answer :
$$\sqrt 3 v$$
Solution :
As the figure drawn above shows that at points $$A$$ and $$B$$ the vertical component of velocity is $$v$$ $$\sin {60^ \circ }$$ but their directions are opposite. Hence, change in velocity.
$$\Delta v = v\sin {60^ \circ } - \left( { - v\sin {{60}^ \circ }} \right) = 2v\sin {60^ \circ } = \sqrt 3 v$$
As the figure drawn above shows that at points $$A$$ and $$B$$ the vertical component of velocity is $$v$$ $$\sin {60^ \circ }$$ but their directions are opposite. Hence, change in velocity.
$$\Delta v = v\sin {60^ \circ } - \left( { - v\sin {{60}^ \circ }} \right) = 2v\sin {60^ \circ } = \sqrt 3 v$$
