An airplane with a speed of 84.5 m/s is climbing upward at an angle of 42.3 ° with respect to the horizontal.When the pl

An airplane with a speed of 84.5 m/s is climbing upward at an angle of 42.3 ° with respect to the horizontal.When the plane's altitude is 794 m,the pilot releases a package.(a) Calculate the distance along the ground,measured from a point directly beneath the point of release,to where the package hits the earth.(b) Relative to the ground,determine the angle of the velocity vector of the package just before impact.
(a)(b)分別是兩個(gè)問題.過程可以及其簡(jiǎn)略但是結(jié)果務(wù)必正確啊.

英語人氣：704 ℃時(shí)間：2020-09-12 12:12:11

優(yōu)質(zhì)解答

落地時(shí)間t,落地的垂直速度為v
v^2-(84.5*sin42.3)^2=2*g*794=15562.4
v=137.1 m/s
t=(v+84.5*sin42.3)/g=19.8 s
1）與投放點(diǎn)的距離=84.5*cos42.3 * t =1237.5 m
2）tanθ=v/(84.5*cos42.3)=2.2
θ=65.5°第一問是求啥??？我對(duì)第一問求的不理解。。。是求飛機(jī) 還是求包裹呢？第一問是求多遠(yuǎn)扔包裹您能不能吧第二問最精確的答案告訴我我算了65.49 65.5 都錯(cuò)了第一問對(duì)了英語還是不行呀！最后一問我開始是不是理解錯(cuò)了？題的意思應(yīng)該是：包開始的速度與落地的速度的夾角？？42.3+（90-65.5）=66.8°

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An airplane with a speed of 84.5 m/s is climbing upward at an angle of 42.3 ° with respect to the horizontal.When the pl

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