Asked by Alc

a grapefruit is tossed straight up with an initial velocity of 50 ft/sec. the grapefruit is 5ft above ground when released. how high does it go before returning to the ground equation? Equation of height in terms of time y= -16t^2+50t+5

Answers

Answered by Steve
y = -16(t - 25/16)^2 + 705/16
so, the vertex is at (25/16,705/16)

That should help.
Answered by ALI NAWAZ
CHUSS
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