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How does the time of flight differ between an object launched from a height parallel to ground and an object launched from the ground with an angle of elevation ?



Sagot :

Answer:

Y = Vy t - 1/2 g t^2       height Y after time t

Y1 = -1/2 g t^2     object launched with no vertical velocity

Y2 = Vy t - 1/2 g t^2   object launched with vertical speed Vy

Obviously Y2 is greater than Y1

Y1 will reach a given negative vertical height sooner than Y2

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