Question

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

Answer 1

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.

Answer 2

you are amazing

Answer 3

I only have 8 fans...

Answer 4

i became a fan but it didnt do anything

Answer 5

my 1st time on this site

Answer 6

Doesn't matter, I don't come here too often either.