Question

Define projectile. Find the relation for Time of flight, maximum height, Horizontal range and maximum horizontal
range when projectile fired at some angle with horizontal.​

Answer 1

★ What is projectile motion ?

Projectile motion is a form of Motion experienced by an object or particle that is thrown near the Earth's surface and moves along a curved path under the action of gravity only .

The graph for projectile motion :

$\setlength{ \unitlength}{30} \begin{picture}(6,6) \put(2,2){ \vector(0,1){5}} \put(2,2){ \vector(1,0){5}} \qbezier(2.1,2)(4.2,9)(6,2) \put(4,2){ \line(0,1){0.3}}\put(4,2.5){ \line(0,1){0.3}}\put(4,3){ \line(0,1){0.3}}\put(4,3.5){ \line(0,1){0.3}}\put(4,4){ \line(0,1){0.3}}\put(4,4.5){ \line(0,1){0.3}}\put(4,5){ \line(0,1){0.3}}\put(4,5.2){ \line(0,1){0.3}}\put(4.1,4){ \bf{ Maximum } }\put(4.1,3.7){ \bf{ height } }\put(3.5,1.5){ \bf{ Range \: (R) } }\put(2.2,6.5){ \bf{ Trajectory} } \qbezier(2.3,2.5)(2.5,2.5)(2.6,2)\put(2.5,2.3){ \bf { \theta } }\put(2.7,4){ \line( - 1, - 1){0.4}}\put(2.7,4){ \line(1, - 2){0.2}}\end{picture}$

Terms related to projectile motion :

$\leadsto \purple{Trajectory}$ - The path of a projectile is called Trajectory.

$\leadsto\purple{Horizontal \:Range}$ - The Horizontal Range of a Projectile is defined as the horizontal displacement of a projectile when the displacement of the projectile in the y-direction is zero.

$\leadsto \purple{Vertical\: Range}$ - Vertical range is the maximum vertical distance a projectile can reach. It is the same as the maximum vertical displacement denoted by (H). It is also known as Maximum height .

★ Important formulas related to projectile motion :

→ The initial velocity of projectile can be resolved into two mutually perpendicular components .

$u_{x} = u cos \theta$

$u_y = u sin \theta$

$\setlength{ \unitlength}{30} \begin{picture}(6,6) \put(2,2){ \vector(0,1){2.5}} \put(2,2){ \vector(1,0){2.5}} \qbezier(2.4,2.3)(2.5,2.5)(3,2) \put(2.1,2){\vector(1,1){2}}\put(2.7,2.3){ \bf { \theta } }\put( 3, 1.5){ \tt{ u \cos \theta } }\put( 1,3 ){ \bf{ u \sin \theta } }\put(4 ,3.9 ){ \bf{ u } }\end{picture}$

$\leadsto T (time\:of\:flight) = \dfrac{(2u sin \theta)}{g}$

$\leadsto R (Horizontal \: range) = \dfrac{u^2 (2 sin\theta cos \theta)}{g}$

$\leadsto H ( Maximum \: height) = \dfrac{u^2 sin^2 \theta}{2g}$

* Note - $\theta = angle \: of \: projection$

$u = initial \: velocity$