1.
Projectile Motion :
For
vertical displacement,
y
= usinθ t - ½gt2……………… (i)
For
horizontal displacement,
X
= ucosθ t
t
= x/ucosθ …………………. (ii)
From
equation (i) and (ii)
y
= usinθ.x/ucosθ – ½g(x/ucosθ).
y
= x tanθ - g/(2u2cos2θ)
is
similar to y = ax + bx2 so the path of projectile is
parabolic.
2.
Time of flight (T) :
T
= 2usinθ/g
Time
of ascent and time of descent are same i.e. t = usinθ/g.
3.
Maximum Height (H) :
H
= u2sin2θ/2g.
4.
Range of Projectile (R) :
R
= u2sin2θ/g.
Rmax
= u2/g.
5.
Maximum height for the projectile with maximum range :
When
a projectile is fired at an angle of 45° then,
Rmax
= u2/g.
For
maximum height
Rmax
= 4H
6.
Two Angle of projection for same horizontal range :
A
projectile is fired from ground with velocity u at an angle of θ =
with horizontal then
Range
(R1) = u2sin2θ/g
For
another angle of projection for same range with same velocity will be
Horizontal
range (R1) = u2sin2θ/g
=
u2/g.sin(180°- 2θ)
=
u2/g.sin{2(90°- θ)}
=
u2/g.sin2θ= R2
θ
and (90°- θ) are the two angle of projection for a projectile with
same range with same velocity.
7.
Velocity and direction of projectile at any height :
The
horizontal component of projectile remain constant through the motion
but vertical component is accelerating. At any point P, at horizontal
displacement y
Horizontal
velocity (Vx) = ucosθ
Vertical
velocity (Vy) is given by
Vy2=
(usinθ)2- 2gy
or,
Vy2= √(u2sin2θ-
2gy)
Resultant velocity
(V) = √(Vx2 + Vy2)
=
√(u2 – 2gy)
∴V
= √(u2 – 2gy)
For
direction α be the direction of resultant with horizontal,
So,
tanθ
= vy/vx = {√(u2sin2θ-2gy)}/{ucosθ}
KE
of projectile of mass m is, KE = ½mv2 = ½m(Vx2+
Vy2).
Some
important tips :
-
In projectile motion acceleration is due to gravity.
-
The path of the projectile is called trajectory.
-
Nature of trajectory is parabolic.
-
Horizontal component of velocity is constant through out the motion.
-
If a projectile is projected so that its range obtained is
maximum.then maximum height attained by it is th of
maximum range.
-
If a person can throw maximum horizontal distance R0, then
he can throw maximum height R0/2.
-
Height is maximum if θ = 90°, Hmax = Rmax.
-
At heighest point angle between velocity and acceleration is 90°.
-
Velocity is minimum at highest point hence kinetic energy is minimum.
-
If a projectile is thrown with speed u at angle θ with horizontal
the projectile makes an angle ‘α’ with horizontal then its
speed, v = ucosθ.Secα.
-
Average velocity during time of ascent (i.e. average velocity when
projectile is at highest point).
Vavg
= u/2.√( 1 + 3cos2θ)
-
Average velocity when projectile strikes to ground,
Vavg.=
ucosθ
-
Change in speed when projectile is at highest point,
Δ
= u(1-cosθ)= 2usin2θ/2
-
If two projectile are projected with same speed at different angles
then for same speed at different angles then for same range,
i.
Sum of angles of projection must be 90°
θ
+ α = 90°,
ii.
Hθ/Hα = tan2θ or cot2α
iii.
(tf)θ/(tf)α = tanθ =
cotα
iv.
R = ½g(tf)θ(tf)α ⇒ R
= ½gt1t2:t1t2 ∝ R
v.
R = 4√( HθHα) ⇒ R2∝ H1H2
vi.
H1+ H2= u2/2g.
Horizontal
projection from Height :
For
vertical motion
y
= ½gt2…………………….(i)
for
horizontal motion
t
= x/u ……………….(ii)
from
equation (i) and (ii),
y
= gx2/2u2
y
= (g/2u2).x2 is the equation of parabola.
For
time of flight :
H
= ½gT2
T
= √(2h/g)
For
horizontal range :
Range
(R) = u.T = u√(2h/g)
Some
important tips :
When
a ball rolled off from top of staircase with horizontal velocity ‘u’
having width ‘b’ and height ‘h’ the ball hits nth
step then, n =
2hu2
gb2
-
If a man hits the target, he should point his gun in a direction
higher than the target.
-
If a man fires his gun directly aimed towards monkey at height, at
same instant monkey at height, at same instant monkey starts falling
then bullet hits the monkey.
-
If two bodies are projected horizontally from certain height with
different velocities u1 and u2 in opposite
direction then
i.
Their velocities are perpendicular after time,
ii.
Velocities of 1st body and 2nd body when their
velocity are perpendicular,v1 =√(u12+
u1u2) and v2 = √(u22+
u1u2)
iii.
Their position vectors are perpendicular of after time,
t
=
2√(u1u2)
g
-
If a ball is droped from height ‘h’ from the top of an inclined
plane of inclination ‘α’, ball elastically collides the it
again, strike the inclined plane at a distance.
S
= 8hSinα.
then the path of the particle is a parabola.
