Derive formula for horizontal range and time taken case of horizontal projection
Answers
Answer:
Trajectory of the projectile
After a time t suppose the body reaches point
P
(
x
,
y
)
then,
Along horizontal axis at
u
x
=
u
(since motion is with uniform horizontal velocity)
a
x
=
0
s
x
=
x
d
i
s
t
a
n
c
e
=
s
p
e
e
d
×
t
i
m
e
or,
x
=
u
×
t
t
=
x
u
(1)
Along vertical axis ,
u
y
=
0
at time
t
=
0
a
y
=
g
s
y
=
y
By second equation of motion,
s
y
=
u
y
t
+
1
2
a
y
t
2
Using equation (1)
and putting
u
y
=
0
at time
t
=
0
y
=
0
+
1
2
g
(
x
u
)
2
y
=
1
2
g
x
2
u
2
y
=
g
x
2
2
u
2
This is the equation of parabola. So, the trajectory of the projectile fired parallel to the horizontal is a parabola.
The velocity of the projectile at any time
Along the horizontal axis,
a
x
=
0
so, velocity remains constant and velocity at
A
along horizontal will also be
u
.
Along vertical,
u
y
=
0
a
y
=
g
By first equation of motion
v
y
=
u
y
+
a
y
t
v
y
=
0
+
g
t
v
y
=
g
t
Magnitude of resultant velocity at any point
P
v
2
=
v
2
x
+
v
2
y
v
=
√
u
2
x
+
g
2
t
2
Now
β
is the angle which resultant velocity makes with the horizontal, then
tan
β
=
v
2
y
v
2
x
=
g
t
u
or,
β
=
tan
−
1
(
g
t
u
)
Time of flight
It is the total time for which the projectile remains in flight (from
O
to
A
). Let
T
be the time of flight. For vertical downward motion of the body we use
s
y
=
u
g
t
+
1
2
a
y
t
2
or,
h
=
0
×
T
+
1
2
g
T
2
or,
T
=
√
2
h
g
Horizontal Range
It is the horizontal distance covered by projectile during the time of flight. It is equal to
O
A
=
R
. So,
R
=
Horizontal velocity
×
Time of flight
=
u
×
T
=
u
√
2
h
g
So,
R
=
u
√