derivetion of tan teta = gt/u
Answers
its a model dear
Consider that the initial velocity of projectile in X and Y direction be v
0x
and v
0y
respectively.
Now, as the particle reaches the point P during its motion, consider v
x
and v
y
be the velocity in X and Y directions respectively.
Using the forst equation of motion for the particle to reach the point P.
In vertical direction:-
v
y
=v
0y
=gt
And in horizontal direction:=
v
x
=v
0x
From the above equation, we can obtain the value of θ
tanθ=
v
x
v
y
=
v
0x
(v
0y
−gt)
θ=tan
−1
v
0x
(v
0y
−gt)
(b)
The maximum height of projectile is given by,
h
m
=
2g
u
0
2
sin
2
θ
...(i)
The horizontal range of the projectile is given by,
R=
g
u
0
2
sin2θ
... (ii)
Taking the ratio of the maximum height and range of projectile:
R
h
m
=
2sin
2
2θ
sin2θ
=
2×2sinθcosθ
Sinθ×sinθ
=
4Cosθ
sinθ
=
4
tanθ
tanθ=(4h
m
/R)
θ=tan
−1
(4h
m
/R)
thank u dear...
taataa....
Explanation:
Consider that the initial velocity of projectile in X and Y direction be v
0x
and v
0y
respectively.
Now, as the particle reaches the point P during its motion, consider v
x
and v
y
be the velocity in X and Y directions respectively.
Using the forst equation of motion for the particle to reach the point P.
In vertical direction:-
v
y
=v
0y
=gt
And in horizontal direction:=
v
x
=v
0x
From the above equation, we can obtain the value of θ
tanθ=
v
x
v
y
=
v
0x
(v
0y
−gt)
θ=tan
−1
v
0x
(v
0y
−gt)
(b)
The maximum height of projectile is given by,
h
m
=
2g
u
0
2
sin
2
θ
...(i)
The horizontal range of the projectile is given by,
R=
g
u
0
2
sin2θ
... (ii)
Taking the ratio of the maximum height and range of projectile:
R
h
m
=
2sin
2
2θ
sin2θ
=
2×2sinθcosθ
Sinθ×sinθ
=
4Cosθ
sinθ
=
4
tanθ
tanθ=(4h
m
/R)
θ=tan
−1
(4h
m
/R)