Question #89895
An object allowed to fall from height of 70m at the same time a projectile is fired straight up at velocity of 35m/s. When and where will they pass each other?
1
Expert's answer
2019-05-17T11:55:36-0400
Y=70(g×t2)/2Y=70- (g\times t^2)/2

is the equation of motion of the object


Y=35×t(g×t2)/2Y=35\times t-(g \times t^2)/2

is the equation of motion of the projectile

Equate these equations:


70(g×t2)/2=35×t(g×t2)/270- (g\times t^2)/2 =35\times t-(g \times t^2)/2

From here: t=2sec; Y=50 meter


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