,
, 3 . 2 | Motion in a Plane
u
u
u
(a) (b) (c)
u
u
u
(d) (e) (f)
Figure 3.1
However, we can separately discuss the motion of the projectile for both the horizontal and vertical parts. In this
regard, we begin by considering the origin as the point of projection.
Now, we have ux = ucos θ; ax = 0 ; uy =
usin θ; ay =
−g.
A
u sin
u
X
O u cos B
Figure 3.2
3.1.1 Horizontal Motion
1
As As a x = 0, we have q and x = ux t + a x t2 = ux t = ut cos q
n x = ux + a x t = ux = ucosand
2
3.1.2 Vertical Motion
In the downward direction, we know that the acceleration of the particle is g. Thus, α y =−g.
Further, the y-component of the initial velocity is uy. Thus,
1
ν y = uy − gt and y = uy t − gt2 ; also we have, ν2y = u2y − 2gy.
2
3.1.3 Time of Flight
Let us suppose that the particle is at B at time t. Therefore, the equation for horizontal motion gives OB = x = ut
cos θ .
However, the y-coordinate at the point B is zero. Thus, from the equation of vertical motion,
1 1 1
=y ut sin θ − gt2 or, =0 ut sin θ − gt2 or, t(usin θ − =gt) 0
2 2 2
2usin θ
Thus, either t = 0 or, t =
g
, 3 . 2 | Motion in a Plane
u
u
u
(a) (b) (c)
u
u
u
(d) (e) (f)
Figure 3.1
However, we can separately discuss the motion of the projectile for both the horizontal and vertical parts. In this
regard, we begin by considering the origin as the point of projection.
Now, we have ux = ucos θ; ax = 0 ; uy =
usin θ; ay =
−g.
A
u sin
u
X
O u cos B
Figure 3.2
3.1.1 Horizontal Motion
1
As As a x = 0, we have q and x = ux t + a x t2 = ux t = ut cos q
n x = ux + a x t = ux = ucosand
2
3.1.2 Vertical Motion
In the downward direction, we know that the acceleration of the particle is g. Thus, α y =−g.
Further, the y-component of the initial velocity is uy. Thus,
1
ν y = uy − gt and y = uy t − gt2 ; also we have, ν2y = u2y − 2gy.
2
3.1.3 Time of Flight
Let us suppose that the particle is at B at time t. Therefore, the equation for horizontal motion gives OB = x = ut
cos θ .
However, the y-coordinate at the point B is zero. Thus, from the equation of vertical motion,
1 1 1
=y ut sin θ − gt2 or, =0 ut sin θ − gt2 or, t(usin θ − =gt) 0
2 2 2
2usin θ
Thus, either t = 0 or, t =
g
