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Answer 1/3
Author:
montgomeryjacobson
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3Given,
u = 30 m/s
angle, tan = 1/8
To find,
At t = 1 sec find measure of angle a
Solution,
tanx = 1/8
sinx = 1/√65
cosx = 8/√65
Along direction of motion after t = 1 sec
v = u + at
vsina = usinx + (-g)t
vsina = 30 × 1/√65 - 10
vsina = 30/√65 - 10
vsina = (30 - 10√65)/√65 (i)
Horizontal component will remain constant throughout the journey
vcosa = ucosx
vcosa = 30 × 8/√65 (ii)
Dividing equation i and ii
[tex]\frac{vsina}{vcosa} = \frac{(30 - 10\sqrt{65})}{\sqrt{65}} \\[/tex] × [tex]\frac{\sqrt{65} }{240}[/tex]
sina/cosa = (30 - 10√65) / 240
tana = (30 - 10√65) / 240
a = tan⁻¹ (30 - 10√65) / 240
Angle a is given as tan⁻¹ (30 - 10√65) / 240.
Author:
frostystevenson
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