define uniform motion what will be value of maximum horizontal range
Answers
Answer:
AVERAGE VELOCITY OF PROJECTILE - DEFINITION
concept
The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with u making an angle θ with the vertical.
There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum =0
It is the horizontal velocity which is uniform and hence v
av
=u
x
=ucosθ
For a general point:
Displacement in Y-direction:
y=usinθ×t−
2
gt
2
Displacement in X-direction:
x=ucosθ×t
Now in order to calculate average velocity:
Average Velocity =
Totaltime
NetDisplacement
RANGE OF A PROJECTILE MOTION - FORMULA
concept
The horizontal distance travelled by a projectile from its initial position (x=y=0) to the position where it passes y=0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight T
f
. therefore, the range R is
R=(v
o
cosθ
o
)(T
f
)=(v
o
cosθ
o
)(2v
o
sinθ
o
)/g ......(1)
Or, R=
g
v
o
2
sin2θ
o
......(2)
Equation 2 shows that for a given projectile velocity v
o
, R is maximum when sin2θ
o
is maximum, i.e. when θ
o
=45
o
.
The maximum horizontal range is, therefore
R
m
=
g
v
0
2
TIME OF FLIGHT OF A PROJECTILE - DIAGRAM
concept
Let, time taken to reach maximum height =t
m
Now, v
x
=v
o
cosθ
o
and v
y
=v
o
sinθ
o
−gt
Since, at this point, v
y
=0, we have:
v
o
sinθ
o
−gt
m
=0
Or, t
m
=(v
o
sinθ
o
)/g
Therefore, time of flight =T
f
=2t
m
−2(v
o
sinθ
o
)/g because of symmetry of the parabolic path.
MAXIMUM HEIGHT OF A PROJECTILE - DIAGRAM
concept
x=v
x
t=(v
o
cosθ
o
)t
and y=(v
o
sinθ
o
)t−(1/2)gt
2
The maximum height h
m
is given by:
y=h
m
=(v
o
sinθ
o
)(
g
v
o
sinθ
o
)−
2
g
(
g
v
o
sinθ
o
)
2
(for t=t
m
)
Or, h
m
=
2g
(v
o
sinθ
o
)
2
Explanation:
Read formulas, definitions, laws from Basics of Projectile Motion here. ... Average Velocity of Projectile - definition ...