A Tennis ball is thrown into Swimming pool from the top of a tall hotel. The height of the ball from the pool is noted by 0(t) - 16t2-4t +300 ft where t is the time in seconds after the ball was thrown. How long after the ball thrown was it 144 feet above the pool?
Answers
Answer:
a tennis ball is thrown into a swimming pool from the top of a hotel.
The height of the ball above the pool is modeled D(t)=-16t^2-4t+20 feet,
where t is the time, in seconds, after the ball was thrown.
How long after the ball is thrown is it at 44 feet above the pool?
:
There is a big problem with this problem:
:
Look at the given equation; -16t^2 - 4t + 20; the 3rd term (20) indicates the
initial height of the ball is only 20 ft, another words the ball was thrown
downward at 4 ft/sec from 20 ft. It would never be at 44 ft obviously.
:
We can assume the top of the hotel is not 20 but 200 ft; when will it be at 44 ft?
-16t^2 - 4t + 200 = 44
-16t^2 - 4t + 200 - 44 = 0
-16t^2 - 4t + 156
Simplify and change the signs, divide by 4
4t^2 + t - 39 = 0
This will factor to:
(4t + 13)(t - 3) = 0
The positive solution is what we want here
t = 3 seconds to descend to 44 ft
:
We can illustrate this graphically, height is vertical, time is horizontal