Architecture
건축 Architecture


Position of a projected object - Time required to reach the highest position

작성자 Uploader : deyoon 작성일 Upload Date: 2016-04-15변경일 Update Date: 2016-04-15조회수 View : 240

Height(Veritcal position) of an object which is projected from at an arbitrary height y0, with its inital velocity V at an angle of θ with respect to the horizontal direction:  

y-direction velocity : Vy=Vsinθ  
x-direction velocity : Vx=Vcosθ  

travel distance in y-direction : Sy(t) = Vy*t-(1/2)*g*t^2 = V*sinθ*t-(1/2)*g*t^2  

Height of the object with respect to time : H(t) = y0 + V*sinθ*t - (1/2)*g*t^2  
Horizontal travel distance of the object with respect to time : Sx(t)=V*cosθ*t  

Time required to fall back to y0 : t2 = 2Vsinθ/g, since V*sinθ*t-(1/2)*g*t^2 = 0  
Time required to reach the highest position : t1 = t2/2, t1=Vsinθ/g  

Time required to reach the ground(y=0) : t3 = (1/g)*(V*sinθ+((V*sinθ)^2 + 2*g*y0)^(1/2)), since H(t) = y0 + V*sinθ*t - (1/2)*g*t^2 = 0  



*** 참고문헌[References] ***

No reference
t1=V*sin(θ)/g
변수명 Variable 변수값 Value 변 수 설 명 Description of the variable




★ 로그인 후 수식작성 및 즐겨찾기에 추가할 수 있습니다.
★ To make new formula or to add this formula in your bookmark, log on please.


코멘트

댓글 입력