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

Hint: Projectile motion is the point at which an object moves in a reciprocally balanced, explanatory way. The way that the object follows is called its trajectory.

Complete answer:
Projectile motion possibly happens when there is one force applied toward the start, after which the main effect on the trajectory is that of gravity.
A projectile is thrown with the velocity v at an angle θ with horizontal.
then, \[{{v}_{x}}=vcos\theta ,{{v}_{y}}=vcos\theta \]
we have to find the vertical component of velocity of the projectile at height equal to half of the maximum height.
we know, \[{{H}_{\max }}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
here u is the initial velocity of projectile but here we have given initial velocity is v
so, \[{{H}_{\max }}=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\]
now, the time taken to reach half of maximum height, t
\[Y={{v}_{y}}t+\dfrac{1}{2}{{a}_{y}}{{t}^{2}}\]
\[\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}=v\sin \theta .t-5{{t}^{2}}\]
\[200{{t}^{2}}-40v\sin \theta .t+{{v}^{2}}{{\sin }^{2}}\theta =0\]
\[t=\dfrac{(\sqrt{2}\pm 1)v\sin \theta }{10\sqrt{2}}\]
\[{{v}_{y}}'={{v}_{y}}+{{a}_{y}}t\]
\[=v\sin \theta -\dfrac{(\sqrt{2}-1)v\sin \theta }{\sqrt{2}}\]
\[=\dfrac{v\sin \theta }{\sqrt{2}}\]
A projectile is any object whereupon the main force is gravity, Projectiles travel with an illustrative trajectory because of the impact of gravity, There are no even forces following up on projectiles and in this manner no level increasing speed, The even velocity of a projectile is consistent (a never showing signs of change in esteem),
There is a vertical speeding up brought about by gravity; it’s worth is 9.8 m/s/s, down, the vertical velocity of a projectile changes by 9.8 m/s each second, the even motion of a projectile is autonomous of its vertical motion.

Note: In projectile motion, horizontal motion and the vertical motion are free of one another; that is, neither one of the motions influences the other. This is the guideline of compound motion set up and utilized by him to demonstrate the explanatory type of projectile motion. The horizontal and vertical segments of a projectile's velocity are autonomous of one another.