Derieve the equation of motion such that an object having a horizontal motion
Answers
Explanation:
An object is thrown with a velocity v
o
with an angle θ with horizontal x-axis.
The velocity component along X-axis= V
ox
=V
o
cosθ
The velocity component along Y-axis =v
oy
=v
0
sinθ
At time T=0, no displacement along X-axis and Y-axis.
So, x
o
=0,y
0
=0
At time T=t,
Displacement along X-axis= v
ox
.t=v
0
cosθ----- (1)
Displacement along Y-axis=v
oy
.t=(v
0
sinθ)t−(
2
1
)gt
2
−−−−−−−(2)
Time required for maximum height,
At maximum height, the velocity component along Y-axis is zero. ie;
v
y
=0 Let the time required to reach maximum height is t
max
Therefore, the initial velocity for motion along Y-axis
v
y
=v
0
sinθ−gt
max
=>0=v
0
sinθ−gt
max
=>t
max
=(v
0
sinθ)/g------- (5)
Total time of flight,
To reach maximum height, time required is v
o
sinθ/g. It takes equal time to reach back ground.
So, total time of flight, T
max
=
g
2(v
0
sinθ)
------- (6)
Maximum height reach,]
Let the maximum height reached by the object be H
max
When body of projectile reaches the maximum height, then
v
y
2
=(v
0
sinθ)
2
=2gH
max
=>0=(v
0
sinθ)
2
=2gH
max
∴H
max
=
2g
(v
0
sinθ)
2
---------- (7)
Horizontal range,
Let R is the horizontal range by the projected body.
Here using horizontal component of velocity only as effective velocity to transverse horizontal path.
R=(v
0
cosθ).T
max
=
g
(v
0
cosθ).2(v
0
sinθ)
=v
0
2
sin2θ/g
R=
g
v
0
2
sin2θ
-------- (8)