VECTOR AND TWO DIMENSION MOTION
R = P 2 + Q 2 + 2PQ cos θ
Q sin θ
tan β =
p + Q cos θ
Q
R
θ
P
RIVER PROBLEM
SHORTEST DISTANCE
VMAN sin θ = VRIVER
VRESULT = VMAN cos θ
β
LEAST TIME
2
2
VRESULT = VMAN
+ VRIVER
y VRIVER
=
D VMAN
PROJECTILE
Whenever acceleration and velocity have any angle except 00 and 1800
Velocity is resolve in horizontal and vertical directions
horizontal component does not change as there is no acceleration
Vertical component changes due to acceleration due to gravity
If body is thrown horizontally it is half projectile (angle with g is 900)
HALF PROJECTILE
2H
T=
g
(x,y)
2H
R = uT = u
g
v( t ) = ui + gtj ⇒ v = u 2 + (gt ) 2
1 x
y = g 
2 v
gt
2
Initial velocity V at an angle of θ with horizontal
H horizontal Range
T total time of flight
R Range
(x,y) any point on trajectory
2V sin θ
T=
g
T=
2u y
g
V 2 sin 2 θ
H=
2g
H=
u 2y
2g
V 2 sin 2θ
R=
g
R=
1
x2
y = x tan θ − g 2
2 V cos 2 θ
2u x u y
g
ESSENTIAL FEATURES OF PROJECTITILE (OBLIQUE)
For maximum Horizontal Range angle with horizontal is 450
For maximum Height angle with horizontal is 900
At complementary angles Range is same
Angle for Range and height to be equal is tan−1(4) = 760
Horizontal velocity at any instant u x = V sin θ
Vertical velocity at any instant u y = V sin θ − gt
Speed at any time is (Vcosθ) 2 + (V sin θ − gt ) 2
Horizontal displacement at any instant = V cos θ.t
1 2
Vertical displacemennt at any time = Vsinθ.t − gt
2