If R is the range of a projectile on a horizontal plane and h its
maximum height, the maximum horizontal range with the
same velocity of projection is: a) 2h
(b) R² / 8h
(c) 2R+h2 /8R. (d) 2h + R2/8h
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Answer:
Explanation:
R=
g
16sin2θ
h=
2g
16sin
2
θ
R
h
=
g
16sin
2
θ
×
16sinθ
g
=
4
1
tanθ
=tanθ=
R
4h
X=
R
2
+(4h)
2
sinθ=
R
2
+16h
2
4h
R
max
=
16h
2
2h(R
2
+16h
2
)
R
max
=
8h
R
2
+2h
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