Question

hola guys !!


please derive the maximum height attended by a projectile!!

thank u !! :-)

have a wonderful day!! :-)​

Answer 1

Projectile Motion is a 2-D motion, and any 2D motion problem can be solved as two 1-D motion problem. In fact, any n-D motion problem can be solved as n 1-D motion problem.

Considering Vertical Motion ;

Since,

$v = u + at \\ = > 0 = u \sin( \theta) - gt \\ = > t = \frac{u \sin( \theta) }{g} = time \: \: taken \: \: to \: \: reach \: \: the \: \: maximum \: \: height \: \: of \: \: the\: \: projectile$

And,

$s = ut + \frac{1}{2} at^{2} \\ = > h = u \sin( \theta) \times \frac{u \sin( \theta) }{g} - \frac{1}{2} \times g \times ( \frac{u \sin( \theta) }{g} )^{2} \\ = > h = \frac{u^{2} \sin^{2} ( \theta) }{g} - \frac{ u^{2} \sin ^{2} ( \theta) }{2g} \\ = > h = \frac{u^{2} \sin^{2} ( \theta) }{2g} = maximum \: \: height \: \: attained \: \: by \: \: the \: \: projectile$

You can also use the 3rd equation of motion and reach the answer very quickly.

Hope, it'll help you.....