Question

Derive experience for maximum height and range for a projectile when projected at angle THETA to the horizontal direction ?​

Answer 1

__________________________________

Maximum Height

It is the maximum vertical height attained by object.

Take Motion from A to H

Take,

• uy = usinθ
• ay = -g
• y0 = 0
• y = h
• t = T/2 = usinθ/g

Use formula :

$\begin{gathered} {\purple{ \sf{y \: = \: y_{0} \: + \: u_{y} t \: + \: \dfrac{1}{2} a_y {t}^{2} }}} \\ \\ \implies \sf{h \: = \: 0 \: + \: u \sin \theta \times \dfrac{ u \sin \theta}{g} \: + \: \frac{1}{2} ( - g) \bigg( \frac{u \sin \theta}{g} } \bigg)^{2} \\ \\ \implies \sf{h \: = \: \frac{ {u}^{2}}{g} \sin^{2} \theta \: - \: \frac{1}{2} \frac{ {u}^{2} \sin^{2} \theta }{g} } \\ \\ \implies \sf{h \: = \: \frac{ {u}^{2} { \sin}^{2} \theta }{2g} }\end{gathered}$

__________________________________

Horizontal Range:

It is the horizontal distance covered by an object.

• u = ucosθ
• t = T
• a = g = 0
• s = R

$\begin{gathered}\green{\sf{s \: = \: ut \: + \: \dfrac{1}{2}gt^2}} \\ \\ \implies {\sf{R \: = \: u \cos \theta T \: + \: 0}} \\ \\ \implies {\sf{R \: = \: u \cos \theta T}} \\ \\ \implies {\sf{R \: = \: u \cos \theta \: \times \: \dfrac{2u \sin \theta}{g}}} \\ \\ \implies {\sf{R \: = \: \dfrac{u^2}{g}2 \sin \theta \cos \theta }} \\ \\ \implies {\sf{R \: = \: \dfrac{u^2 \sin 2 \theta}{g}}}\end{gathered}$

__________________________________

Time of Flight:

It is the total time

Motion from A to H and H to B

• uy = usinθ
• ay = -g
• t = T/2
• vy = 0

Time of flight = Time of Ascent + Descent Time

$\sf \: T = t + t \\ \sf \: » t = T/2$

Now use :

$\begin{gathered}{\red{\sf{v_y \: = \: u_y \: + \: a_yt}}} \\ \\ \implies {\sf{0 \: = \: u \sin \theta \: + \: (-g) \dfrac{T}{2}}} \\ \\ \implies {\sf{T \: = \: \dfrac{2u \sin \theta}{g}}}\end{gathered}$

__________________________________

$\sf{@WildCat7083 }$