Consider an individual drop of water coming out of a decorative water fountain 3 feet above ground level with a velocity of 30 ft/sec.

Question

Consider an individual drop of water coming out of a decorative water fountain 3 feet above ground level with a velocity of 30 ft/sec. The path of the water can be modeled by the equation, y = -16t 2 + v0t + h0, where v0 is the initial velocity and h0 is the initial height. When will the height of the water droplet be back at the initial height of the fountain

Answer 1

Hey Kishi!

Here, we're given a function and we're trying to find when y (the height) will be at 3 (the initial height), so let's plug in our values!

y = -16t^2 + v0t + h0

3 = -16t^2 + 30t + 3

0 = -16t^2 + 30t

16t^2 -30t = 0

t(16t - 30) = 0

t = 0 and 16t-30= 0

16t - 30 = 0

16t = 30

t = 15/8

So we know that the height is at 3 feet at 0 seconds and 15/8 seconds, but we rule out 0 seconds because that's the initial height. Therefore, our answer would be 15/8 seconds