Question

The velocity of the projectile when it is at height equal to half of the maximum height is
A. $\upsilon \sqrt{{{\cos }^{2}}\theta +\dfrac{{{\sin }^{2}}\theta }{2}}$
B. $\sqrt{2}\ \upsilon \cos \theta$
C. $\sqrt{2}\ \upsilon \sin \theta$
D. $\upsilon \ \tan \theta \ \sec \theta$

Answer 1

Hint In this type of question,solve separately the velocity of x-direction and y-direction and then use the formula to find the projectile velocity.
Now, we would put the value of projectile height in the formula for the velocity in y-direction.After that we find the x and y component of the the velocity of projection.Further, calculate $V_y$ and $V_x$ at the maximum height separately. Then solve the V for $V_y$ and $V_x$ because $V=\sqrt{V_{y}^{2}+V_{x}^{2}}$ and hence we will get our final result.

Complete Step-by-Step solution
First we will see motion in y-direction is in perpendicular direction.
We know the formula of maximum height
$H=\dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{2g}$ (Here u is initial velocity and is the projection angle)
For velocity in y-direction
$v_{y}^{2}=u_{y}^{2}-2gh$
Where uy is initial velocity in y direction g is acceleration due to gravity
$\therefore h=\dfrac{1}{2}\dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{2g}$
$\therefore \upsilon _{y}^{2}$ is
$\upsilon _{y}^{2}=u_{y}^{2}-2g\left( \dfrac{1}{2}\times \left( \dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{2g} \right) \right)$
$\upsilon _{y}^{2}=u_{y}^{2}-2g\left( \dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{4g} \right)$
$\upsilon _{y}^{2}-{{\upsilon }^{2}}{{\sin }^{2}}\theta =-2g\dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{4g}$
Here $\upsilon _{y}^{2}$ is the y component of the projection vector u.
$\therefore {{u}_{y}}=\upsilon \sin \theta$.
$\upsilon _{y}^{2}={{\upsilon }^{2}}{{\sin }^{2}}\theta -\dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{2}$
$\upsilon _{y}^{2}=\dfrac{{{\upsilon }^{2}}{{\sin }^{2}}\theta }{2}$
$\upsilon _{y}^{2}\dfrac{\upsilon \sin \theta }{\sqrt{2}}$ (I)
Also, motion in horizontal director or x-direction and it is given by
${{\upsilon }_{x}}=\upsilon \cos \theta$ (II)
as the Vx is the x component of the projection vector so it is .

& \therefore v=\sqrt{v_{y}^{2}+v_{x}^{2}} \\\ & \ \ \ \ \ =\sqrt{\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2}+{{v}^{2}}{{\cos }^{2}}\theta } \\\ \end{aligned}$$ taking $v^2$ as common $$=\sqrt{{{v}^{2}}\left( \dfrac{{{\sin }^{2}}\theta }{2}+{{\cos }^{2}}\theta \right)}$$ $=v\sqrt{{{\cos }^{2}}\theta +\dfrac{1}{2}{{\sin }^{2}}\theta }$ $\therefore v=v\sqrt{{{\cos }^{2}}\theta +\dfrac{1}{2}{{\sin }^{2}}\theta }$ is the correct answer. the velocity at half of maximum height is $v\sqrt{{{\cos }^{2}}\theta +\dfrac{{{\sin }^{2}}\theta }{2}}$ **Note** The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion. If any object is thrown with the velocity u, making an angle Θ from horizontal, then the horizontal component of initial velocity = u cos Θ and the vertical component of initial velocity = u sin Θ. The horizontal component of velocity (u cos Θ) remains the same during the whole journey as herein, no acceleration is acting horizontally.