Question

a projectile is thrown with velocity v at an angle theta with horizontal. when the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

Answer 1

a projectile is thrown with velocity v at an angle $\bf{\theta}$ with horizontal.
then, $\bf{v_x=vcos\theta}$
and $\bf{v_y=vcos\theta}$

A/c to question,
we have to find vertical component of velocity of the projectile at height equal to half of the maximum height.
we know, $\bf{H_{max}=\frac{u^2sin^2\theta}{2g}}$
here u is intial velocity of projectile. but here given initial velocity is v
so, $\bf{H_{max}=\frac{v^2sin^2\theta}{2g}}$

now, time taken to reach half of maximum height, t
$\bf{y=v_yt+\frac{1}{2}a_yt^2}$

$\bf{\frac{v^2sin^2\theta}{4g}=vsin\theta.t-5t^2}$
$\bf{200t^2-40vsin\theta.t+v^2sin^2\theta=0}$
$t=\bf{\frac{40vsin\theta\pm20\sqrt{2}vsin\theta}{400}}$
$t=\bf{\frac{(\sqrt{2}\pm1)vsin\theta}{10\sqrt{2}}}$
[ I just take '-' sign , you can take '+' sign too]

now, vertical component of velocity ,
$\bf{v'_y=v_y+a_yt}\\=vsin\theta-g\times\frac{(\sqrt{2}-1)vsin\theta}{10\sqrt{2}}\\=vsin\theta-\frac{(\sqrt{2}-1)vsin\theta}{\sqrt{2}}\\=\frac{vsin\theta}{\sqrt{2}}$

Answer 2

Answer:

just look at abhi's answer

Explanation:

H max = $\frac{u^{2}sintheta}{2g}$

from second equation of motion considering only y component

s = ut + 1/2a$t^{2}$

$\frac{u^{2}sintheta}{2g}$ = (u sintheta)(t) + 1/2 × -10 × $t^{2}$

200$t^{2}$ - 40 (u sintheta) t + $(u sintheta)^{2}$

using this to find t we input and find u to be (usintheta)/2