Question

A projectile is fired with a velocity u making an angle theta with the horizontal. Derive an expression for a) maximum height b) time of flight c) horizontal range d)maximum horizontal range

Answer 1

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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 :

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

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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

T = t + t

» t = T/2

Now use :

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

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Horizontal Range :

It is the horizontal distance covered by an object.

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

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

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Maximum Horizontal Range :

Take θ = 45°

$\implies {\sf{R \: = \: \dfrac{u^2}{g} \sin 2 \: \times \: 45}} \\ \\ \implies {\sf{R_{max} \: = \: \dfrac{u^2}{g}}}$

Answer 2

$\huge{\sf{Projectile }}$

A parabolic projectilon making a maximum angle of 45° with respect to the horizontal frame

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}$

The initial velocity of the projectile is u and angle with the x - axis is ∅

• Velocity of a projectlie along X axis is u cos∅

• Velocity along Y - axis is u sin∅

Maximum Height

• The final velocity vector along y axis tends to zero

1. Maximum distance covered along y axis

From,

$\sf \: { {v}^{2} }_{y} - { {u}^{2} }_{y} = 2( - g)h \\ \\ \longrightarrow \: \sf{ - u {}^{2} \sin( \theta) {}^{2} = - 2gh} \\ \\ \longrightarrow \: \sf \: 2h = \frac{ {u}^{2} \sin {}^{2} ( \theta) }{g} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf{h = \dfrac{ {u}^{2} {sin}^{2} \theta }{g} }}}$

Time of Flight

• Total time interval which involves the time of ascent and time of descent of a projectile

$\sf T_f = t_a + t_d$

Consider s = ut + 1/2at²

Since,

The projectile starts from the ground and returns to the ground,the vertical displacement is zero » s = 0 m

$\sf{0 = uT + \dfrac{1}{2}( - g) {T}^{2} } \\ \\ \longrightarrow \: \sf \: uT = \dfrac{1}{2} g {T}^{2} \\ \\ \longrightarrow \: \sf \: 2u \sin( \theta) = gT \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf{T = \dfrac{2u \sin( \theta) }{g}}}}$

Time of ascent and Time of Descent would be u sin∅/g

Range

• Horizontal displacement of a projectlie (along X axis )

$\sf{speed = \dfrac{distance}{time} } \\ \\ \longmapsto \: \sf{u \cos( \theta) \times \dfrac{2u \: \sin( \theta) }{g} = R} \\ \\ \longmapsto \: \sf \: R = \dfrac{ {u}^{2} }{g} \times 2 \sin( \theta) \cos( \theta) \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{R = \dfrac{ {u}^{2} \sin(2 \theta) }{g} }}}$

For maximum range,∅ = 45°

$\longmapsto \: \boxed{ \boxed{ \sf { R = \dfrac{ {u}^{2} }{g} }}}$