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Standard XII
Physics
The Equation for the Path of Projectile
40 the equati...
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40 the equation of trajectory of a projectile projected with a velocity of 10m/satanangle45degreewithverticalisgivenby(g=10m/s)(a)y^2=10(x+y)(b)y^2=10(x-y)(c)x^2=10(x+y)(d)x^2=10(x-y)
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Q.
The equation of trajectory of a projectile is
y
=
10
x
−
⎧
⎪ ⎪ ⎪
⎩
5
9
⎫
⎪ ⎪ ⎪
⎭
x
2
If we assume
g
=
10
m
s
−
2
, the range of projectile (in metres) is:
Q.
The equation of trajectory of a projectile is
y
=
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x
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(
5
9
)
x
2
. If we assume
g
=
10
m
s
−
2
the range of projectile ( in metres) is:
Q.
Equation of trajectory of a projectile is given by
y
=
−
x
2
+
10
x
where x and y are in meters and x is along horizontal and y is vertically upward and particle is projected from origin. Then which of the following options is/are correct.
(
g
=
10
m
/
s
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)
.
Q.
The equation of trajectory of a projectile is
y
=
10
x
−
[
5
9
]
x
2
,
(
m
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. The maximum height is reached when
x
is
Q.
The equation of a projectile is
y
=
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10
x
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4
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m
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The Equation for the Path of Projectile
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