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 b

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 b
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.

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 b
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°

斜抛运动，分解为
水平方向x：匀速直线运动，初速度V0x=84.5*Cos42.3,位移Sx=V0x*t
竖直方向y：自由落体运动,初速度v0y=84.5*Sin42.3向上,位移Sy=Vy*t-0.5gt^2
飞机扔包，落地时间t就是到达最高时间t1+下落过程t2：t1=voy/g,上升了距离Sy=V0y*t1-0.5gt1^2,整个下降距离sy+H=0.5gt2^2从而...

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斜抛运动，分解为
水平方向x：匀速直线运动，初速度V0x=84.5*Cos42.3,位移Sx=V0x*t
竖直方向y：自由落体运动,初速度v0y=84.5*Sin42.3向上,位移Sy=Vy*t-0.5gt^2
飞机扔包，落地时间t就是到达最高时间t1+下落过程t2：t1=voy/g,上升了距离Sy=V0y*t1-0.5gt1^2,整个下降距离sy+H=0.5gt2^2从而知t2
水平距离就是Sx=vox*(t1+t2)->解（1）
竖直方向：落地竖直方向速度Vty=gt2,落地与水平成角度a=arctan(vty/vox)->解（2）
t1=5.8,sy=165, t2=14, t=19.8, sx=1237.5
vty=-137.2,a=artan2.2

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