Question

Please answer fast. ​

Answer 1

Solution :-

Now the initial velocity is given = v

And the angle between projectile and horizon = ∅

Now we have to find out the vertical component of the velocity at H = h/2 .

The vertical component of the velocity

= v sin∅ let it be m

Then as we know :-

v² - u² = -2gH ( taking upward motion as positive)

→ v² - m² = -2gH

Now at H = max , v = 0

→ v² - m² = -2gH

→ 0 - m² = -2gH

→m² = 2gH

$H = \dfrac{m^2}{2g}$

or At H = h/2

$H' = \dfrac{m^2}{4g}$

Now by applying

→ v² - u² = -2gH again.

→v² = m² - 2gH'

$\rightarrow v = \sqrt{ m^2 -2.g \dfrac{m^2}{4g} }$

$\rightarrow v = \sqrt{ m^2 - \dfrac{m^2}{2} }$

$\rightarrow v = \sqrt{ \dfrac{2m^2 - m^2}{2} }$

$\rightarrow v = \sqrt{ \dfrac{m^2}{2} }$

$\rightarrow v = \dfrac{m}{\sqrt{2}}$

$\rightarrow v = \dfrac{v.Sin\theta}{\sqrt{2}}$

Answer 2

Answer:

the correct answer is option b

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