Question

Derivation of H= u square sin squaretheta divided by 2 g

Answer 1

${\bold{\mathfrak{Explantion:}}}$

${\bold{\huge{\underline{SOLUTION-}}}}$

• In given question information given about eqn of projectile.

• We have to derive that eqn.

$\underline \bold{derivation : } \\ \implies h_{max} = \frac{ {u}^{2} \sin^{2} \theta }{2g}$

• According to given question :

$\bold{1. \: neglect \: air \: resistance} \\ \bold{2. \: g \: constant} \\ \\ \bold{in \: y \: direction : } \\ \implies u_{y} = u \sin \theta \\ \implies a_{y} = g \\ \implies s = dy = h_{max} \\ \bold{by \: second \: eqn \: of \: motion} \\ \implies {v}^{2} = {u}^{2} + 2as \\ \implies {v}^{2} _{y} = {u}^{2} + 2ay.dy \\ \implies {0}^{2} = ({usin \theta})^{2} + 2( - g) h_{max} \\ \implies - {u}^{2} \sin^{2} \theta = - 2g \times h_{max} \\ \implies {u}^{2} { \sin }^{2} \theta = 2g \times h_{max} \\ \bold{\implies h_{max} = \frac{ {u}^{2} { sin }^{2} \theta }{2g} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {\bold{derived}}$