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)分别是两个问题.过程可以及其简略 但是结果务必正确啊.
落地时间t,落地的垂直速度为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）与投放点的距离=84.5*cos42.3 * t =1237.5 m
2）tanθ=v/(84.5*cos42.3)=2.2
θ=65.5°第一问是求啥啊？ 我对第一问求的不理解。。。是求飞机 还是求 包裹呢？第一问是求 多远扔包裹您能不能吧第二问最精确的答案 告诉我我算了65.49 65.5 都错了第一问对了英语还是不行呀！最后一问我开始是不是理解错了？题的意思应该是：包开始的速度与落地的速度的夹角？？42.3+（90-65.5）=66.8°