Consider a projectile, say a ball, thrown horizontally with an initial velocity `vecu` from the top of a tower of height h.
As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity u and a vertical downward distance because of constant acceleration due to gravity g. Thus, under the combined effect the ball moves along the path OPA. The motion is in a 2-dimensional plane. Let the ball take time t to reach the ground at point A. Then the horizontal distance travelled by the ball is x(t) = x, and the vertical distance travelled is y(t)=y.
We can apply the kinematic equations along the x direction and y direction separately. Since this is two-dimensional motion, the velocity will have both horizontal component `u_(x)` and vertical component `u_(y)`.
Motion along horizontal direction: The particle has zero acceleration along direction So, the initial velocitly `u_(x)` remains constant throughout the motion.
The distance travelled by the projectile at a time is given by the equation `x=u_(x)t+(1)/(2)at^(2)`.
Since a=0 along x direction we have
`x=u_(x)t" ".......(1)`
Motion along downward direction: Here `u_(y)=0` (initial velocity has no downward component), a s (we choose the +ve waxis in downward direction), and distance y at time t.
`:.` From equation, `y=u_(x)t+(1)/(2)at^(2)` we, get
`y=(1)/(2)at^(2)" "......(2)`
Substituting the value of from equation (i) in equation (ii) we have
`y=(1)/(2)g""(x^(2))/(u_(z)^(2))=((g)/(2u_(x)^(2)))`
`y=Kx^(2)" ".......(3)`
where `K=(g)/(2u_(x)^(2))` is constant.
Equation (3) is the equation of a parabola. Thus, the path followed by the projectile is a parabola.
1. Time of Flight: The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight.
Consider the example of a tower and projectile. Let h be the height of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.
We know that `s_(y)=u_(y)t+(1)/(2)at^(2)` for vertical motion. Here `s_(y)=h,t=T,u_(y)=0` (i.e., no initial vertical velocity). Then
`h=(1)/(2)"gt"^(2)orT=sqrt((2h)/(g))`
Thus, the time of flight for projectile motion depends on the height of the tower, but is independent of the horizontal velocity of projection. If one ball falls vertically and another ball is projected horizontally with some velocity, both the balls will reach the bottom at the same time. This is illustrated in the figure.
Horizontal range: The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range. For horizontal motion, we have
`s_(x)=u_(x)t+(1)/(2)at^(2)`
Here, `s_(x)=R` (range), `u_(x)=u,a=0` (no horizontal acceleration) T is time of flight. Then horizontal range =uT.
Since the time of flight `T=sqrt((2h)/(g))`,we substitute this and we get the horizontal range of the particle as `R=usqrt((2h)/(g))`.
The above equation implies that the range R is directly proportional to the initial velocity u and inversely proportional to acceleration due to gravity g.
3. Resultant Velocity (Velocity of projectile at any time): At any instant 1, the projectile has velocity components along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instanti, as shown in figure.
The velocity component at any talong horizontal (x-axis) is `v_(x)=u_(x)+a_(x)t.`
Since, `u_(x)=u,a_(x)=0,` we get
`v_(x)=u`
The component of velocity along vertical direction (y-axis) is `v_(y)=u_(y)+a_(y)t`
Since, `u_(y)=0,a_(y)=g,` we get
`v_(y)="gt"`
Hence the velocity of the particle at any instant is
`vecv=uhati+"gt"hatj`
The speed of the particle at any instant is given by
`v=sqrt(v_(x)^(2)+v_(y)^(2))`
`:.v=sqrt(u^(2)+g^(2)t^(2))`
Speed of the projectile when it hits the ground: When the projectile hits the ground after initially thrown horizontally from the top of tower of. heighth, the time of thight is
`t=sqrt((2h)/(g))`
The horizontal component velocity of the projectile remains the same i.e. ` v_(x)=u`.
The vertical component velocity of the projectile at time T is
`v=gT=gsqrt((2h)/(g))=sqrt(2gh)`
The speed of the particle when it reaches the ground is
`v=sqrt(u^(2)+2gh)`
