Question

derice the equation of maximum height for projectile motion ​

Answer 1

Answer:

hope the answer will help you ☺️☺️..

Explanation:

An object launched into projectile motion will have an initial launch angle anywhere from 0 to 90 degrees. ... The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g .

Answer 2

EXPLANATION.

$\sf : \implies \: derive \: the \: equation \: of \: maximum \: height \: for \: projectile \: motion$

$\sf : \implies \: { \underline{in \: y \: direction}} \\ \\ \sf : \implies \: u_{y} = u \sin( \theta) \\ \\ \sf : \implies \: a_{y} = - g \: = for \: a \to \: b \: \\ \\ \sf : \implies \: v_{y} = 0 \\ \\ \sf : \implies \: from \: newton \: 1st \: equation \: of \: kinematics$

$\sf : \implies \: v_{y} = u_{y} + a_{y}t \\ \\ \sf : \implies \: 0 = (u \sin \theta) - gt \\ \\ \sf : \implies \: (u \sin \theta) = gt \\ \\ \sf : \implies \: t \: = \frac{u \sin( \theta) }{g}$

$\sf : \implies \: from \: newton \: 3rd \: equation \: of \: motion \\ \\ \sf : \implies \: v {}^{2} _{y} = u {}^{2}_{y} + 2a_{y}d_{y} \\ \\ \sf : \implies \: {0}^{2} - (u \sin \theta) {}^{2} = 2( - g) h_{m} \\ \\ \sf : \implies \: h_{m} = \frac{u {}^{2} { \sin {}^{2} ( \theta) }^{} }{2g}$

$\sf : \implies \: { \underline{in \: x \: direction}} \\ \\ \sf : \implies \: u_{x} = u \cos\theta \\ \\ \sf : \implies \: a_{x} = 0 \\ \\ \sf : \implies \: t \: = \frac{2u \sin( \theta) }{g} \\ \\ \sf : \implies \: d_{x} = r = u_{x}.t \\ \\ \sf : \implies \: r \: = u \cos( \theta) \times \frac{2u \sin( \theta) }{g} \\ \\ \sf : \implies \: r \: = \frac{ {u}^{2} \sin( 2\theta) }{g}$

$\sf : \implies \: { \underline{additional \: information}} \\ \\ \sf : \implies \: if \: we \: move \: directly \: from \: a \to \: c \: \\ \\ \sf : \implies \: u_{y} = u \sin( \theta) \\ \\ \sf : \implies \: a_{y} = - g \\ \\ \sf : \implies \: d_{y} = 0 \\ \\ \sf : \implies \: from \: newton \: second \: equation \: of \: kinematics$

$\sf : \implies \: s \: = ut + \dfrac{1}{2} a {t}^{2} \\ \\ \sf : \implies \: 0 = u \sin( \theta)t - \frac{1}{2} g {t}^{2} \\ \\ \sf : \implies \: t \: = \frac{2u \sin( \theta) }{g}$