college algebra...gravity formula?

You throw it up at a certain speed, it falls down and you're concerned about when it's 10 meters above the water.

I wouldn't call it college algebra, it was something we did in grade school.

College Algebra Formulas

The general equation for the height of an object as a function of time is:

h(t) = h0 + (v0)t - (g)t²

Where h0 is the initial height, v0 is the initial velocity and g is the acceleration due to gravity.

You are given that h0 = 44 m and v0 = 11 m/s

For Earth, g = 9.8 m/s²

This makes your specific equation become.

h(t) = 44 m + (11 m/s)t - (9.8 m/s²)t²

You are asked for the time when the height is 10 m so you substitute 10 m for h(t):

10 m = 44 m + (11 m/s)t - (9.8 m/s²)t²

Subtract 10 m from both sides:

0 = 34 m + (11 m/s)t - (9.8 m/s²)t²

Multiply both sides by -1 and rearrange things a bit:

(9.8 m/s²)t² - (11 m/s)t - 34 m = 0

Because the time cannot be negative, use only the positive value of the quadratic formula:

t = {11 m/s + √[(-11 m/s)² - 4(9.8 m/s²)(-34 m)]}/{2(9.8 m/s²}

t = 2.5 s

There are two parts. You should know Galilean formulae well enough.

In each part, acceleration, a = g = 9.78m/s². u = 11 m/s.

For the upwards motion, to know its maximum height when its final speed, v = 0, use

v² - u² = -2gh,

0 - 11² = -2(9.78)h,

+11² =+2(9.78)h,

h = 121/19.56 = 6.1860m.

Time taken, is got by using

(v -u)/t = -g,

t = (v -u)/(-g) = u/g

= 11/9.78 = 1.1247 s

In the next part (downward motion) the total height it has to go through is

6.1860+44 -10 = 40.186m.

Use

h = ut+½ gt² {since u = 0 here}

h = ½ gt²

40.186 = 4.89t²,

t = √(40.186/4.89) = √8.2180 s = 2.8667

Total time = 1.1247 + 2.8667 =3.9914 s