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

Question

anonymous · Answer

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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