Question

一道英文数学题目(追加10分)

In priates of the caribbean, a cannon was fired at an enemy ship. The muzzle velocity was 60m/s. At what angle above the horizontal must the cannon have been aimed in order to hit the enemy ship 320 m away?(Dont worry about intial height on this one. The ships will be at the same level due to calm seas)
我用算式y=-4.9t^2+60sin(theta)t 和x=60 cos(theta)t 然后设320=60 cos(theta)t 然后我知道要用sin theta/cos theta 但是怎么做呢?

Answer 1

你的式子我没看懂？
有一个隐含条件：自由落体速度为9.81（或10）米每秒平方
其实就是炮弹的纵向速度（60sinx)每秒减（9.81）10的情况下直到其为负的60sinx，这时其横向移动距离，及（60cosx）乘以（120sinx）除以9.81（或10）的商的积是320
中间用一下（sinx）平方加（cosx）平方为1的转换公式

追问

我用的是y=-(1/2)g^2*t+Vo(sin theta)*t+ho和x=Vo(cos theta)t的公式.
然后我得到y=-4.9t^2+60(sin theta)t和x=60cos(theta)t. 然后因为可以设x=320m. 就变成了320=60cos(theta)t.
那之后要怎么求呢?

追答

你的y是什么？

追问

y是horizontal motion, x是vertical motion啊

Answer 2

吧