Question

drive the equation for maximum height​

Answer 1

AnswEr:

The maximum height, ymax, can be found from the equation:

$\\ \sf{vy^2 = voy^2 + 2 a_{y} (y - y_{o}) }$

yo = 0, and, when the projectile is at the maximum height, $\sf{v_{y}= 0.}$

Solving the equation for ymax gives:

$\\ \sf{y_{max} = {- v_{oy}}^2 /(2a_{y})}$

Plugging in voy = vo sin(q) and ay = -g, gives:

$\sf{y_{max} = {v_{o}}^2sin^2 (q) /(2 g)}$

where g = 9.8 m/s²

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