AboutAbe Mantell Expertise Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
Question I'm trying to help some students with quadratic application probs. The one that I'm not sure how to complete is:
A person throws a water balloon out of an office building that is 15 ft high and tries to land it in a bucket 39 ft away. the path of this balloon is illustrated by the equation y = -(1/16)x^2 + 2x + 15. If the bucket has a diamter of 8 ft and is exactly 39 ft away, will it land in the bucket?
I got the vertex ( 16 sec, 31 ft) but I'm not sure how to figure out the horizontal distance from building to bucket. I want to use the distance formula but not sure if I use the points (0, 15) and (16, 15) which would I think gives me the horizontal distance to the vertex and then double that. Hope this makes sense and appreciate any help-thanks a senior citizen trying!
Answer Hello Alan,
Assuming the bucket is at ground level, with y=0, we just need to
determine the x-value of where it will land. So, solve
-(1/16)x^2 + 2x + 15 = 0 gives: x= 16-4*sqrt(31) or 16+4*sqrt(31)
Taking the positive value, x=16+4*sqrt(31) which is about 38.27 ft
from the building. So, the balloon will definitely land in the bucket
(that's a BIG bucket, 8 ft diameter!!!).
