A field gun is pointed at an angle of 35 degree above the horizontal and fires a shell with a muzzle velocity of 650 m/s. Ignoring air resistance , to what peak height will the shell rise and what will be its range ??? Please answer !!!
Answers
Answer:
The distance is
1.11
m
.
Explanation:
To answer at the question we have to find the angle
α
above the horizontal, at which we have to shoot.
The motion is a parabolic motion, that is the composition of two motion:
the first, horizontal, is an uniform motion with law:
x
=
x
0
+
v
0
x
t
and the second is a decelerated motion with law:
y
=
y
0
+
v
0
y
t
+
1
2
g
t
2
,
where:
(
x
,
y
)
is the position at the time
t
;
(
x
0
,
y
0
)
is the initial position;
(
v
0
x
,
v
0
y
)
are the components of the initial velocity, that are, for the trigonometry laws:
v
0
x
=
v
0
cos
α
v
0
y
=
v
0
sin
α
(
α
is the angle that the vector velocity forms with the horizontal);
t
is time;
g
is gravity acceleration.
To obtain the equation of the motion, a parabola, we have to solve the system between the two equation written above.
x
=
x
0
+
v
0
x
t
y
=
y
0
+
v
0
y
t
+
1
2
g
t
2
.
Let's find
t
from the first equation and let's substitue in the second:
t
=
x
−
x
0
v
0
x
y
=
y
0
+
v
0
y
x
−
x
0
v
0
x
−
1
2
g
⋅
(
x
−
x
0
)
2
v
2
0
x
or:
y
=
y
0
+
v
0
sin
α
x
−
x
0
v
0
cos
α
−
1
2
g
⋅
(
x
−
x
0
)
2
v
2
0
cos
2
α
or
y
=
y
0
+
sin
α
x
−
x
0
cos
α
−
1
2
g
⋅
(
x
−
x
0
)
2
v
2
0
cos
2
α
To find the range we can assume:
(
x
0
,
y
0
)
is the origin
(
0
,
0
)
, and the point in which it falls has coordinates:
(
0
,
x
)
(
x
is the range!), so:
0
=
0
+
sin
α
⋅
x
−
0
cos
α
−
1
2
g
(
x
−
0
)
2
v
2
0
cos
2
α
⇒
x
⋅
sin
α
cos
α
−
g
2
v
2
0
cos
2
α
x
2
=
0
⇒
x
(
sin
α
cos
α
−
g
2
v
2
0
cos
2
α
x
)
=
0
x
=
0
is one solution (the initial point!)
x
=
2
sin
α
cos
α
v
2
0
g
=
v
2
0
sin
2
α
g
(using the double-angle formula of sinus).
Now let's find, finally
α
.
sin
2
α
=
g
x
v
2
0
=
9.8
⋅
120
250
2
=
0.0188
°
⇒
2
α
1
=
arcsin
0.0188
⇒
2
α
=
1
,
078
°
⇒
α
=
0.53
°
and
2
α
2
=
180
°
−
arcsin
0.0188
=
178.92
°
⇒
α
=
89.47
°
.
The solution are both valid, but the most reasonable is the first.
Using trigonometry, now, we can find the answer.
There is a right-angled triangle with an acute angle of
0.53
°
, and a cathetus of
120
m
. The other cathetus is
c
=
120
⋅
tan
0.53
=
1.11
m
.
Hope it helps and please make me brainlist