A projectile is fired with a velocityuma king an angle(teta)with the horizontal.Show that its trajectory is a parabola.Also find expression forits(1)Maximum height and(2).Horizontal range
Answers
Answer:
ANSWER
Consider the following equations of a projectile with angle
of projection θ and initial velocity v
0
- here
x=v
ox
t refer to book ;
rearrange the expression for time t ,
t=
v
0x
x
substituted
v
ox
x
for in the expression
y=(v
0
sinθ)t−
2
1
gt
2
y=v
oy
(
v
ox
x
)−
2
1
g(
v
0x
x
)
2
substitute v
o
cosθ for v
ox
sin θ for v
oy
in the above expression.
Y=(v
0
sinθ)(
v
o
cosθ
x
)−
2
1
g(
v
o
cosθ
x
)
2
Hence the obtained equation of the projectile is
Y=(tanθ)x−(
2(v
o
)
2
gsec
2
θ
)x
2
The expression is to be obtained in the form of
Y = ax+bx
2
.
hence it is a parabolic motion so from it we get all the required quantities
( 1 ) T =
g
2v
o
sinθ
( 2 ) H
max
=
2g
v
o
2
sin
2
θ
( 3 ) Range =
g
v
o
2
sin2θ
solution