A projectile is fired with a velocity u making an angle theta with the

1.	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» $\huge{\sf{Projectile }}$ A parabolic projectilon MAKING a maximum ANGLE of 45° with respect to the horizontal FRAME $\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\<klux>VECTOR</klux>(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){$\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 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° $\longmapsto \: \boxed{ \boxed{ \sf { R = \dfrac{ {u}^{2} }{g} }}}$

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»

$\huge{\sf{Projectile }}$

A parabolic projectilon MAKING a maximum ANGLE of 45° with respect to the horizontal FRAME

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\<klux>VECTOR</klux>(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){$\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

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°

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