Find the ratio of maximum height attained by two projectiles thrown with initial velocity 40° and 50 °
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Explanation:
Range of projective =R= g 4 sin2θ
u= initial velocity, θ= angle of projection
V A =2v,v B =v,θ B
=45 o
R B =
g
v
2
sin90
o
=R
B
=
g
v
2
Now
R
A
=
g
4v
2
sin2θ
for both ranges equal
R
A
=R
B
g
4v
2
sin2θ
a
=
g
v
2
sin2θ
A
=
4
1
θ
A
=
2
1
sin
−1
(
4
1
)
A will be projected at an angle of
θ=
2
a
sin
−1
(
4
1
)
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