when a ball is dropped off a 50 foot building, it is h feet above the ground after t seconds where t= sqrt(50-h)/4 how long for the ball to hit ground how far above the ground will the ball be after 1 second

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when a ball is dropped off a 50 foot building, it is h feet above the ground after t seconds where t= sqrt(50-h)/4 how long for the ball to hit ground how far above the ground will the ball be after 1 second

Mathematics
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t= sqrt(50-h)/4 When it hits the ground, h=0 so: t= sqrt(50)/4 = 1.767767 seconds. after one second, t=1, so 1 = sqrt (50-h) /4 sqrt (50-h) = 4 50 -h = 16 h=34 feet.
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