Question

answer it............. ​

Answer 1

✡ EXPLANATION ✡

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

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

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

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

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

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

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

Answer 2

Answer:

Let a projectile be projected at an angle θ

$h \: = \: \frac{ {u}^{2 \ \: } {sin}^{2} θ}{g} - \frac{ {u}^{2} {sin}^{2} θ \: }{2g}$

with the initial velocity u.

From the diagram, we get,

Uₓ = x/t

⇒x = Uₓ × t

Substitute the values in the above equation.

x = u cos θ

$⇒t = \frac{x}{u \: cos \:θ \: }$

Maximum height

The height reached by the body when projected vertically upwards where the vertical velocity is zero.

Using the law of motion equation we will further continue to find the expression of maximum height.

H = Uyt + ½ gt²

Further solve the above equation

$⇒ \: h \: = u \: sin \: θ \: ( \frac{u \: sin \:θ \: }{g} ) - \frac{1}{2} g \: \\ ( \frac{u \: sin \: θ}{g} {)}^{2}$

$h \: = \: \frac{ {u}^{2} {sin}^{2} θ \: }{2g}$

hope this helps you