Question

A projectile is thrown with velocity V at an angle with the horizontal . When the projectile is at height equal to half of the maximum height then the vertical component of the velocity of projectile is

Answer 1

Question :

A projectile is thrown with velocity V at an angle with the horizontal . When the projectile is at height equal to half of the maximum height then the vertical component of the velocity of projectile is ?

Theory :

Maximum Height Reached in Projectile motion:

${\boxed{\sf{\green{H=\frac{u^2\:sin^2\theta}{2g}}}}}$

Solution :

Given : velocity of projectile = v

angle of projection = θ

initial velocity has two components

$v_{x} = v \cos( \theta) \: and \: v _{y} = v \sin( \theta)$

so after time t vertical component of velocity:

$u _{y} = v _{y} - gt = v \sin( \theta) - gt$

_________________________

we have to find vertical component of velocity of the projectile at height equal to half of the maximum height.

$\sf{h=\frac{u^2\:sin^2\theta}{4g}}$

Use equation of motions

$h = v _{y}t + \frac{1}{2} gt {}^{2}$

$\frac{v {}^{2} \sin( \theta) }{4g} = v \sin( \theta) - 5t {}^{2}$

$\implies \: v {}^{2} \sin( \theta) = 40v \sin( \theta)t - 200t {}^{2}$

$\implies \: 200t {}^{2} - 40v \sin( \theta) + v {}^{2} \sin( \theta) = 0$

$\implies \: t = \frac{40v \sin( \theta) \frac{ + }{} 20 \sqrt{2} v \: \sin( \theta) }{10 \sqrt{2} }$

$\implies \: t = \frac{( \sqrt{2} \frac{ + }{} 1)v \sin( \theta) }{10 \sqrt{2} }$

Now vertical component of velocity:

$u _{y} = v _{y} - gt$

$u_{y} = v \sin( \theta) - g \times \frac{( \sqrt{2} - 1)v \sin( \theta) }{10 \sqrt{2} }$

$u_{y} = v \sin( \theta) - \frac{( \sqrt{2} - 1)v \sin( \theta) }{ \sqrt{2} }$

$u _{y}= \frac{v \sin( \theta) }{ \sqrt{2} }$

which is the required solution!