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The time when an object that is thrown upright vertically meets an object that is thrown vertically downward from above

ÀÛ¼ºÀÚ Uploader : anyway ÀÛ¼ºÀÏ Upload Date: 2019-06-24º¯°æÀÏ Update Date: 2019-10-21Á¶È¸¼ö View : 612

Calculation of the time 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)8*g*dt^2

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


*** Âü°í¹®Çå[References] ***

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