How do you solve for "t" for this problem?

Question

s = 1/2 g t^2 + v0t < --- > (distance an object falls)


t = ???

Answer 1

Ah, a little general physics in the math section...

s = 1/2 g t^2 + v0t

rearrange to get a more familiar form...

1/2 g t^2 + vo t - s = 0

now you have a quadratic equation in t that can be solved using the usual methods.

Answer 2

1.0 If initial velocity is zero or its a free fall then initial velocity
Vo=0, therefore Vot = 0, so you will just work with s=1/2gt^2,

by cross multiplication, t^2=2s/g

then, t = Square root of (2s/g)

there are two possible values of t:

t1= +square root of (2s/g) and,
t2= -square root of (2s/g) ,

disregard the negative value for time.

where:

s= height of fall
g= 9.81 meter per second squared

2.0 If there is an initial velocity, Vo is not zero, then transform the equation to follow the quadratic equation format as follows:

Quadratic Equation : AX^2 + BX + C = 0

for s = 1/2gt^2 + Vot, deduct both side with s,

it will now become, s-s = 1/2gt^2 + Vot - s,
or 0=1/2gt^2 + Vot - s, or simply

1/2gt^2 + Vot - s = 0: from Quadratic equation therefore,

A = 1/2g
B = Vo
C = - s
X = t

Solution using Quadratic formula:

X= {-B +/- Square root of [B^2-4AC]}/2A

substituting; t = {-Vo +/- Square Root [Vo^2 - 4*1/2g*(-s)}/(2*1/2g)
t={-Vo +/- Square Root [Vo^2 + 2gs]}/g

+/- means that t has two possible values:

t1 = {-Vo + Square Root [Vo^2 + 2gs]}/g, and
t2 = {-Vo - Square Root [Vo^2 + 2gs]}/g

I hope this helps.

Answer 3

You solve for t by recognizing that your equation is a quadratic equation in t. Rearrange terms, then solve for t using the famous: minus b plus or minus the square root of b squared minus 4ac all over 2a.

Answer 4

sqrt(2s/g) = t