Question

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

Answer 1

${ Here is your answer }$

▶${ Intial velocity u = v sina }$
${ top height v = 0 }$
${ v² - u² = -2gh }$
${ 0- (v sina)²= -2gh }$
${ h = (v sina)² ÷ 2g }$
${ h' = h ÷ 2 }$
${ h' = (v sina)² ÷ 4g }$

▶${ v'² - u² = -2gh' }$
${ v' = √( v sina )² - 2g ( v sina )² ÷ 4g }$
${ v' = v sina ÷ √2 }$

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