# you throw a stone straight down. the stone starts with a velocity of 54 feet per second down. it leaves the sling 326 feet above the ground?

i need the equation for algebra

### 5 Answers

- Andrew SmithLv 77 months ago
Adding a question mark does not make a question.

ASKING a question makes a question.

What did you hope to find?

AN equation could be as simple as

h = 326

or

v = -54

These are both equations.

as is h = vt + 1/2 g t^2

or v^2 = u^2 + 2gh

v = u + gt

All of these are equations that MIGHT be relevant.

And I could think of so many more.

To ask a question you need to have some clear idea of what it is that you need to find.

- electron1Lv 77 months ago
There are two numbers that you can find. They are the stone’s velocity just before it hits the ground and time the stone is moving. The acceleration is 32 ft/s^2. Use the following equation to determine the stone’s velocity just before it hits the ground.

vf^2 = vi^2 + 2 * a * d

vf^2 = 54^2 + 2 * 32 * 326

vf = √23,780

This is approximately 154 ft/s. Use the following equation to determine the time.

vf = vi + a * t

√23,780 = 54 + 32 * t

t = (√23,780 – 54) ÷ 32

This is approximately 3.13 seconds. If you use a slightly different number for the acceleration, your answers will be slightly different. I hope this is helpful for you.

- LônLv 77 months ago
It depends on what you are trying to find...i)the final velocity (v) or ii) the time it takes to reach the ground (t)

i) v^2 = u^2 + 2aS

ii) S = ut + at^2/2

- oldprofLv 77 months ago
The stone's height above impact point is given by y(t) = h - Ut - 1/2 gt^2; where h = 326 ft, U = 54 fps, and g = 32.2 ft/s^2. To solve for t = ? time to impact, we set y(t) = 0 and solve the quadratic eqn.

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