Question

A projectile is fired with a velocity u, making an angle θ with the horizontal. Show that its trajectory is a parabola and find out its time of flight , maximum height attained and horizontal range.

Answer 1

2D PROJECTILE DERIVATIONS:

Along Y axis :

$y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2} \: \: \: \: . \: . \: . \: .(1)$

Along X axis :

$x = u \cos( \theta) t \: \: \: \: . \: . \: . \: .(2)$

Now, put value of t in eq.(1) :

$\implies y = \dfrac{u \sin( \theta) x }{u \cos( \theta) } - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2} ( \theta)}$

$\boxed{ \implies y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2} ( \theta)} }$

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For time of flight, put y = 0 (as y axis displacement is zero):

$0 = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$\boxed{\implies \: t = \dfrac{2u \sin( \theta) }{g} }$

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For Range :

$x = u \cos( \theta) \times t$

$\implies \: x = u \cos( \theta) \times \dfrac{2u \sin( \theta) }{g}$

$\boxed{ \implies \: x = \dfrac{{u}^{2} \sin(2 \theta) }{g} }$

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For max height :

$h= u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$\implies \: h= u \sin( \theta) \bigg \{ \dfrac{u \sin( \theta) }{g} \bigg \} - \dfrac{1}{2} g { \bigg \{ \dfrac{u \sin( \theta) }{g} \bigg \}}^{2}$

$\boxed{ \implies \: h = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g} }$

Hope It Helps.