Question

The vertical component of the velocity of the projectile at the half of maximum height is:
A) $3v{\text{ }}\sin \theta$
B) $v{\text{ }}\sin \theta$
C) $\dfrac{{v{\text{ }}\sin \theta }}{{\sqrt 2 }}$
D) $\dfrac{{v{\text{ }}\sin \theta }}{{\sqrt 3 }}$

Answer 1

An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independence of the horizontal and the vertical components of projectile motion in his Dialogue on the Great World Systems (1632). The horizontal distance travelled by a projectile from its initial position to the position where it passes y = 0 during its fall is called the horizontal range, R.

Complete step by step solution:
Initial speed of the projectile = $v$
Angle of projectile = $\theta$
Vertical component of the initial velocity = $vsin \theta$
Horizontal component of the initial velocity = $vcos \theta$
Horizontal component of the final velocity = $vcos \theta$
At maximum height
$\Rightarrow {v_f} = 0$
By using third equation of motion, we get:
$\Rightarrow {v^2} - {u^2} = 2as$
$\Rightarrow {(0)^2} - {(vsin \theta )^2} = - 2gh$
$\therefore h = \dfrac{{{{(v{\text{ }}\sin \theta )}^2}}}{{2g}}$
$\Rightarrow {h_1} = \dfrac{h}{2}$
$\Rightarrow {h_1} = \dfrac{{{{(v{\text{ }}\sin \theta )}^2}}}{{4g}}$
By using the second equation of motion, we get;
$\Rightarrow {h_1} = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow \dfrac{{{{(v{\text{ }}\sin \theta )}^2}}}{{4g}} = v{\text{ }}\sin \theta \times t + \dfrac{1}{2}( - 10){t^2}$
$\Rightarrow 200{\text{ }}{t^2} - 40{\text{ }}r\sin \theta \times t + {v^2}{\sin ^2}\theta = 0$
$\Rightarrow t = \dfrac{{(\sqrt 2 - 1)v{\text{ }}\sin \theta }}{{10\sqrt 2 }}$
Vertical component of the velocity of the projectile is:
$\Rightarrow {v_y} = u + at$
$\Rightarrow {v_y} = v{\text{ }}\sin \theta - g\dfrac{{(\sqrt 2 - 1)v{\text{ }}\sin \theta }}{{10\sqrt 2 }}$
On solving the further equation, we get:
$\Rightarrow {v_y} = \dfrac{{v{\text{ }}\sin \theta }}{{\sqrt 2 }}.$

Therefore, option(C) is correct.

Note: An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity.