Question

a projectile is projected with velocity v making an angle theta with horizontal direction, find its horizontal range
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Answer 1

Answer:

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Answer 2

Answer:

${\huge{\boxed{\mathcal{\green{answer :-}}}}}$

The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

${v}^{2} = {u}^{2} - 2as$

$0 = {u}^{2} { \sin }^{2} θ - 2gh$

${\huge{\boxed{\mathcal{\red{\: h = \frac{ {u}^{2} { \sin }^{2}θ }{2g} }}}}}$

Explanation:

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