A projectile is fired in such a way that the horizontal range is three times the maximum height.Angle of projection is
Answers
Answer:
R=3H
R=4H cot(theta)
so now, 3H = 4H cot(theta)
cot(theta) = 3/4
tan(theta)= 4/3 = perpendicular /base
So, angle of projection is 53 degree.
Explanation:
I've done full derivations here. If you're not interested I've put coloured titles above the important results that I've used.
So the projectile is fired with some initial velocity
v
0
at some angle
θ
.
This means that the initial velocity in the y direction is given by
v
y
0
=
v
0
sin
(
θ
)
and the initial velocity in the x direction is given by
v
x
0
=
v
0
cos
(
θ
)
.
We now use Newton's 2nd law in the two independent directions. Define upwards as the positive y direction and motion to the right as the positive x direction. Let's consider the x direction first:
m
d
2
x
d
t
2
=
0
Integrating gives
d
x
d
t
=
C
=
v
x
0
Displacement in x as function of time
⇒
x
(
t
)
=
v
x
0
t
=
v
0
t
cos
(
θ
)
Now look at the y direction:
m
d
2
y
d
t
2
=
−
m
g
d
y
d
t
=
−
g
t
+
C
At
t
=
0
,
d
y
d
t
=
v
y
0
=
v
0
sin
(
θ
)
⇒
d
y
d
t
=
−
g
t
+
v
0
sin
(
θ
)
Integrate again
Displacement in y as function of time
⇒
y
(
t
)
=
−
1
2
g
t
2
+
v
0
t
sin
(
θ
)
Now we have basic equations for the position in each direction as functions of time. From this we need to obtain expressions for the range and the maximum height in order to solve the problem.
To find the range, we need to determine the time of flight. We do this by setting y = 0. This will give us two times, one at
t
=
0
which corresponds to before the projectile is launched and the other which shall be when it lands - the time of flight!
0
=
−
1
2
g
t
2
+
v
0
t
sin
(
θ
)
1
2
g
t
2
−
v
0
t
sin
(
θ
)
=
0
t
(
1
2
g
t
−
v
0
sin
(
θ
)
)
=
0
For this to be true we must have either
t
=
0
or
(
1
2
g
t
−
v
0
sin
(
θ
)
)
=
0
For time of flight (which we shall denote
t
f
) we take the second one:
Time of flight
t
f
=
2
v
0
sin
(
θ
)
g
Now that we have an expression for the time of flight, we can sub it into our formula for x distance to find the range (denoted
d
):
Range
d
=
x
(
t
f
)
=
v
0
cos
(
θ
)
⋅
t
f
=
2
v
2
0
sin
(
θ
)
cos
(
θ
)
g
=
v
2
0
sin
(
2
θ
)
g
Almost there, we just need to find an expression for the maximum height. We do this by finding the critical point of the y value, ie we set
d
y
d
t
=
0
We shall denote the time at which the projectile is at max height as
t
m
, solving we get
Time at max height
t
m
=
v
0
sin
(
θ
)
g
This is exactly half the time taken for the projectile to reach the end (can you see why this makes sense?).
Finding the max height (
h
) is simply a case of plugging in this value, ie
h
=
y
(
t
m
)
=
−
1
2
g
t
2
m
+
v
0
sin
(
θ
)
t
m
h
=
−
1
2
g
(
v
0
sin
(
θ
)
g
)
2
+
v
2
0
sin
2
(
θ
)
g
Max Height
h
=
v
2
0
sin
2
(
θ
)
2
g
And now, finally, we are ready to answer the question. We want to know the value of
θ
such that
d
=
3
h
.
∴
v
2
0
sin
(
2
θ
)
g
=
3
v
2
0
sin
2
(
θ
)
2
g
⇒
sin
(
2
θ
)
=
3
2
sin
2
(
θ
)
2
sin
(
θ
)
cos
(
θ
)
=
3
2
sin
2
(
θ
)
sin
(
θ
)
(
3
2
sin
(
θ
)
−
2
cos
(
θ
)
)
=
0
sin
(
θ
)
=
0
or
tan
(
θ
)
=
4
3
The first gives
θ
=
n
π
which doesn't make much sense so we look at the second:
θ
=
arctan
(
4
3
)
≅
53.13
degrees (there are other solutions to this but this is the first and only one that makes physical sense)