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The height where an object thrown upright vertically meets an object thrown vertically downward from above

ÀÛ¼ºÀÚ Uploader : anyway ÀÛ¼ºÀÏ Upload Date: 2019-06-24º¯°æÀÏ Update Date: 2019-10-21Á¶È¸¼ö View : 302

Calculation of the height at which an object thrown upward at vertical speed Vu (m/s) and an object thrown at speed Vd (m/s) vertically downward from height H (m) after dt (s). 

Height of the object thrown upward (m) 
yu = Vu*t - (1/2)*g*t^2 

Height of the object thrown downward from H (m)
yd = H - Vd*(t-dt) - (1/2)*g*(t-dt)^2
 
Since yu = yd,

Vu*t - (1/2)*g*t^2 = H - Vd*(t-dt) - (1/2)*g*(t-dt)^2

Vu*t + Vd*t - g*dt*t = H +Vd*dt - (1/2)*g*dt^2

t = (H +Vd*dt - (1/2)*g*dt^2)/(Vu+Vd-g*dt)

Thus, the height where two objects meet,

yu = Vu*((H +Vd*dt - (1/2)*g*dt^2)/(Vu+Vd-g*dt)) - (1/2)*g*((H +Vd*dt - (1/2)*g*dt^2)/(Vu+Vd-g*dt))^2

*** Âü°í¹®Çå[References] ***

yu = Vu*((H +Vd*dt - (1/2)*g*dt^2)/(Vu+Vd-g*dt)) - (1/2)*g*((H +Vd*dt - (1/2)*g*dt^2)/(Vu+Vd-g*dt))^2
º¯¼ö¸í Variable º¯¼ö°ª Value º¯ ¼ö ¼³ ¸í Description of the variable


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