Asked by Neal parikh
An arrow has an initial launch speed of 39 m/s. If it must strike a target 39 m away at the same elevation, what should be the projection angle?(Hint: sin2θ = 2sinθcosθ)
Answers
Answered by
Damon
u = 39 cos T
39 = u t = 39 cos T * t
t = 1/cosT total time in air
rise time = t/2 = 1/(2cosT)
vertical problem
0 = 39 sin T - g * 1/(2 cos T)
but g/2 = 4.9 m/s^2
4.9/cos T = 39 sin T
cos T sin T = 4.9/39
2 sin T cos T = 9.8/39
so
sin 2T = .251
2T = 15.55
T = 7.28 degrees
39 = u t = 39 cos T * t
t = 1/cosT total time in air
rise time = t/2 = 1/(2cosT)
vertical problem
0 = 39 sin T - g * 1/(2 cos T)
but g/2 = 4.9 m/s^2
4.9/cos T = 39 sin T
cos T sin T = 4.9/39
2 sin T cos T = 9.8/39
so
sin 2T = .251
2T = 15.55
T = 7.28 degrees
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