  Question

Derive the equations of projectile motion.

Solution

An object is thrown with a velocity $${ v }_{ o }$$ with an angle $$\theta$$ with horizontal x-axis.The velocity component along X-axis= $${ V }_{ ox }={ V }_{ o }cos\theta$$The velocity component along Y-axis =$${ v }_{ oy }={ v }_{ 0 }sin\theta$$At time T=0, no displacement along X-axis and Y-axis.So, $$x_{ o }=0,\quad { y }_{ 0 }=0$$At time T=t,Displacement along X-axis= $${ v }_{ ox }.t={ v }_{ 0 }cos\theta$$----- (1)Displacement along Y-axis=$${ v }_{ oy }.t={ (v }_{ 0 }sin\theta )t-(\dfrac { 1 }{ 2 } )g{ t }^{ 2 }-------\quad (2)$$Time required for maximum height,At maximum height, the velocity component along Y-axis is zero. ie;$${ v }_{ y }=0$$  Let the time required to reach maximum height is $${ t }_{ max }$$Therefore, the initial velocity for motion along Y-axis$${ v }_{ y }={ v }_{ 0 }sin\theta -g{ t }_{ max }\\ =>0={ v }_{ 0 }sin\theta -g{ t }_{ max }\\ =>{ t }_{ max }=({ v }_{ 0 }sin\theta )/g$$------- (5)Total time of flight,To reach maximum height, time required is $${ v }_{ o }sin\theta /g$$. It takes equal time to reach back ground. So, total time of flight, $$T_{ max }=\dfrac { 2({ v }_{ 0 }sin\theta ) }{ g }$$------- (6)Maximum height reach,]Let the maximum height reached by the object be $${ H }_{ max }$$When body of projectile reaches the maximum height, then $${ v }_{ y }^{ 2 }={ ({ v }_{ 0 }sin\theta ) }^{ 2 }=2g{ H }_{ max }\\ =>0=\quad { ({ v }_{ 0 }sin\theta ) }^{ 2 }=2g{ H }_{ max }$$$$\therefore { H }_{ max }=\dfrac { { ({ v }_{ 0 }sin\theta ) }^{ 2 } }{ 2g }$$---------- (7)Horizontal range,Let R is the horizontal range by the projected body.Here using horizontal component of velocity only as effective velocity to transverse horizontal path.$$R=({ v }_{ 0 }cos\theta ).{ T }_{ max }=\dfrac { ({ v }_{ 0 }cos\theta ).2({ v }_{ 0 }sin\theta ) }{ g } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ v }_{ 0 }^{ 2 }sin2\theta /g$$$$R=\dfrac { { v }_{ 0 }^{ 2 }sin2\theta }{ g }$$-------- (8)  Physics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 