A body is projected at an angle θ with the horizontal with a velocity u. Show
that its path is a projectile. Obtain equation for maximum height and horizontal
range.
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11th
Physics
Motion in a Plane
Projectile Motion
A projectile is fixed with ...
PHYSICS
A projectile is fixed with certain velocity u making an angle θ with the horizontal. With necessary diagram prove that the trajectory of the projectile is a parabola and also obtain an expression for
(i)Time of flight
(ii)max height reach
(iii)horizontal range of projectile
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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