A stone $(m = 2 k g)$ is projected in a vertical place at angle $θ = 60^{o}$ with the horizontal, with an initial speed $u = 20 m / s$ . The angular momentum of the stone when it is at the maximum height is? $(g = 10 m / s^{2})$ .

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Solution

Normally in a projectile (influenced only by gravity) only the vertical velocity component is changed by the action of gravity, but it's horizontal component will be a constant.
when its angle from the horizontal is $45°45°$ the horizontal and vertical components must be equal (as proved in the picture).
therefore to find the time when the angle gets to $45 °$ , we can find the time when which the vertical velocity component of the projectile reaches the magnitude of the horizontal velocity component which is a constant in this case.
· using the basic equation $“ v = u + a t ”$ for an object moving with a constant acceleration.(where v:-final velocity, u:- initial velocity, acceleration, t:-time )
· rearranging the equation to get the time taken for the initial velocity to reach the final velocity.
$t = (v - u) / a t = (v - u) / a$
· substituting the values for this case , considering the vertical velocity components in the upwards direction .
$t = (100 c o s 60 ° - 100 s i n 60 °) / (- 10) t = (100 c o s 60 ° - 100 s i n 60 °) / (- 10)$

$= (50 - 50 \sqrt 3) / (- 10) = (50 - 50 \sqrt 3) / (- 10)$

$= - 36.6 / - 10 = - 36.6 / - 10$

$= 3.66 s$

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