22. A ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet. What is the maximum height that the ball will reach? @calclover

Question

22.	A ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet.    What is the maximum height that the ball will reach? @calclover

anonymous · Answer

@undeadknight26

undeadknight26 · Answer

._.
Your making me do other peoples work?!

undeadknight26 · Answer

Whats the quadratic formula @Shorty1234 ?

undeadknight26 · Answer

I g2g rn hun sorry https://answers.yahoo.com/question/index?qid=20100801171808AAX9pVD

undeadknight26 · Answer

@confluxepic this is all you bro

anonymous · Answer

Do you know this formula?
$$s=\frac{ v_{f}^{2}-v_{i}^{2} }{ 2a }$$

anonymous · Answer

What is $$s _{0}$$ I mean what does it mean?

anonymous · Answer

No

anonymous · Answer

hello?

anonymous · Answer

Sorry, having connection issues.

anonymous · Answer

This formula is even simpler:
H = (initial velocity)^2 / (2*g)

anonymous · Answer

H is the maximum height.

anonymous · Answer

I'm wondering, do you get some kind of collection of formulas while doing these kind of problems?

anonymous · Answer

Anyway the formula was derived from one of the kinematic equations.

anonymous · Answer

I didn't get one.

anonymous · Answer

except for the quadratic formula/discriminant

anonymous · Answer

what's g?

anonymous · Answer

@aleroth

anonymous · Answer

g is the acceleration due to gravity, 9.8 m/s^2

anonymous · Answer

how do i find g??

anonymous · Answer

It's well known as 9.8

anonymous · Answer

I dunno, I'm seeing this as a physics problem

anonymous · Answer

im taking pre-calc so....

anonymous · Answer

Here's how I'd solve it:
H = 4^2 / 2*9.8 = 0.8163 ft (using the formula)
10 + 0.8163 = 10.8163 ft     (adding on the 10 we started at)