Question

a boy an angle theta with the horizontal. derive the equation throws a javelin at for time of flight and horizontal range of the projectile ​

Answer 1

Given:

A boy threw a javelin at an angle theta with the horizontal.

To find:

• Time of flight

• Horizontal range

Solution:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,-1){\vector(0,1){5}}\put(-1,0){\vector(1,0){8}}\qbezier(0,0)(3,5)(6,-0)\multiput(3,0)(0,0.38){7}{\qbezier(0,0)(0,0)(0,0.2)}\put(3.4,1){\bf H}\put(3,-0.7){\bf R}\put(2.8,-0.6){\vector(-1,0){2.7}}\put(3.5,-0.6){\vector(1,0){2.6}}\put(0,0){\vector(1,2){1}}\put(1.1,2.3){\bf\large u}\end{picture}$

This is an example of GROUND-GROUND projectile.

• Here the 2D motion will be considered as two simultaneously occuring linear motions along X and Y axis.

So, let initial velocity be u.

Let time of flight be t , such that the Y axis displacement will be zero (javelin starts and ends on ground).

$\sf \therefore \: y = u_{y}t - \dfrac{1}{2} g {t}^{2}$

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

$\sf \implies\: u \sin( \theta) t = \dfrac{1}{2} g {t}^{2}$

$\sf \implies\: u \sin( \theta) = \dfrac{1}{2} g t$

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

Now, the horizontal range can be calculated by multiplying the X axis velocity with the time of flight.

$\sf \therefore \: R = u_{x} \times t$

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

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

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

Hope It Helps.