200分求一道数学题,求懂英文的高手If a projectile is fired from ground level with an initial velocity of ν ft/sec and at an angle of θ degrees with the horizontal,the range R of the projectile is given by the following formula.If ν = 9

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200分求一道数学题,求懂英文的高手Ifaprojectileisfiredfromgroundlevelwithaninitialvelocityofνft/secandatanangleofθde

200分求一道数学题,求懂英文的高手If a projectile is fired from ground level with an initial velocity of ν ft/sec and at an angle of θ degrees with the horizontal,the range R of the projectile is given by the following formula.If ν = 9
200分求一道数学题,求懂英文的高手
If a projectile is fired from ground level with an initial velocity of ν ft/sec and at an angle of θ degrees with the horizontal,the range R of the projectile is given by the following formula.If ν = 90 ft/sec,approximate the angles that result in a range of 140 feet.(Please enter your answer(s) in degrees without units and round to the nearest hundredth.Enter your answers as a comma-separated list.)
R=（V^2/16）sinθcosθ
θ =多少好像答案是有2个解

200分求一道数学题,求懂英文的高手If a projectile is fired from ground level with an initial velocity of ν ft/sec and at an angle of θ degrees with the horizontal,the range R of the projectile is given by the following formula.If ν = 9
R=（V^2/16）sinθcosθ
140=90^2/16*1/2*sin2θ
sin2θ=140*32/8100=448/810=224/405
2θ=arcsin224/405=33.579012963518934(度)
或
2θ=180度-arcsin224/405=180度-33.579012963518934(度)
所以
θ=16.7895064818度约等于=16.79度
或
θ约等于73.21度.

17或73

ν = 90 ft/sec
R=（V^2/16）sinθcosθ =1/32*(90^2)*sin2θ=140
sin2θ=224/405=0.553
2θ=33.57°或146.43°
θ=16.8°或73.2°

140=8100/16sinθcosθ
2sinθcosθ =0.55308641974
sin2θ=0.55308641974
θ=17或73

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