Question

How to find the range of a projectile

Answer 1

Answer:

The range (R) of the projectile is the horizontal distance it travels during the motion. Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence: y = utsina - ½ gt2.

hope it helps you.....

Answer 2

$\setlength{\unitlength}{2.8cm}\begin{picture}(1,1)\qbezier(0,0)(0.65,0.7)(1.3,0)\put(0,0){\vector(1,0){1.5}}\put(0,0){\vector(0,1){1}}\put(-0.1,-0.1){{O}}\put(1.5,-0.2){{X}}\put(-0.1,1){{Y}}\put(0,0){\vector(1,1){0.23}}\put(0.12,0.22){{u}}\put(0.5,-0.1){{u.cosx}}\put(-0.32,0.5){{u.sinx}}.\put(0.5,-0.3){{Range}}\put(0.84,-0.28){\vector(1,0){0.6}}\put(0.45,-0.28){\vector(-1,0){0.53}}\put(1,0.5){{Angle\:with\:horizontal=\:'x'}}\end{picture}$

$\sf{Given}:\\\\\sf{velocity\:vertical\:direction\:= u.sinx}\:\\velocity\:in\:horizontal\:direction=u.cosx\\\\Time\:Period=\dfrac{2usinx}{g}\\\\\:Horizontal\:range=horizontal\:velocity\times\:Time\:Period.\\\\\therefore\:Range=\dfrac{2usinx}{g}\times ucosx\\\\\implies\:\dfrac{u^{2}\times2sinx.cosx}{g}\\\\\implies\:\bold{Range=\dfrac{u^{2}sin2x}{g}}$