Question

A water balloon is tossed in the air with an upward velocity of 25 ft/s. The height h(t) in ft after t seconds is given by the function h(t) = -16t^2+25t+3. Please explain your process.

Answer 1

What do we need to solve for exactly?

Answer 2

How many seconds until the balloon hits the ground...

Answer 3

Ah okay. So the first thing we want to do is understand what this equation defines (excuse me if I belittle you in anyway, I just like to start from the beginning). It relates the amount of time passed with the height of the ball. I'll use derivatives here to make it easier to explain using one formula. If you don't know how to find the derivative or what it is, just reply back asking for an explanation:

The position function must equal 0, so our objective is to find the t that gives us h(t)=0.
As a side note, the "3" at the end of the height function is the starting point, since when t=0, h(0)=3 (3ft).
The derivative of the position equation is the velocity equation:

v(t)=-32t+25

This new velocity equation is simpler, and is directly related to what we want to find. At the max height of the path of the balloon, the velocity will be equal to 0, since it is just then stopping to start falling back down. We can find the t that gives us a velocity of 0 using simple algebra:

v(t)=0=-32t+25
-25=-32t
t=25/32

We now know how long it takes to reach the apex of the path. This is approximately half the total time, but unfortunately we start off at a height different from the landing point. That does tell us, however, that the time required to reach the ground after reaching the apex will be slightly longer, since there is a larger distance to travel. We can isolate the two parts of this dilemma, by realizing that it will take another 25/32 seconds to reach the initial 3ft, and then solving for the rest. We can add the 50/32 (full arc) to the little part we will do now:

First we want to find how fast the balloon is at the 3ft marker using our handy little velocity equation:

v(50/32)=-25 [Since there is no such thing as negative velocity, this is essentially 25 ft/s]

Notice how the speed matches the initial throwing speed, since it will always fall down with the same speed it started with (at the starting point, so this will not be the velocity at the end of our full path). Basically all we want to do now is figure out how long it will take this balloon to travel 3ft when accelerating at 32 ft/s. We can reuse the height function with these new variables:

h(t)=16t^2+25t
Where h(t)=3.

This set-up basically tells you how long it will take an accelerating (since gravity is helping it now) balloon with an initial speed of 25 ft/s (because that's how fast it's going at 3ft from the ground) to reach the ground which is 3ft away. Now all you have to do is use the quadratic formula to solve for t. Hope that helps! Sorry for the extremely long wait on my answer, I got busy with other things in between.

Answer 4

whats the answer

Answer 5

X=1.674