The angle of projection of a projectile which is
projected with certain velocity from ground is π/8
with
horizontal and its horizontal range is R. The angle
with the horizontal for another projectile having same
range R and speed is
(1) 3π/8
(2) π/4
(3) π/3
(4) π/6
Answers
Answered by
7
Answer:
1 st option
Explanation:
R=u²sin2Ф÷g
=u²sin(2*180÷4)÷g
=u²sin(45)÷g
u²×0.707÷g
we known that π =180
by the options put Ф=3π÷8 in formula
R=u²sin2Ф÷g
R=u²sin(2×3π÷8)÷g
R=u²sin(3×π÷4)÷g
R=u²sin135÷g
we know sin135=0.707
R=u²×0.707÷g
Answered by
1
Answer:
3π/8
Explanation:
=∫
8
π
8
3π
sinx+cosx
cosx
dx
∫
0
a
f(x)dx=∫
0
a
f(a+b−x)dx
∫
8
π
8
3π
sinx+cosx
cosx
dx=∫
8
π
8
3π
cosx+sinx
sinx
dx=I
2I=∫
8
π
8
3π
⋅1⋅dx
2I=
4
π
I=
8
π
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