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Tigeree
2006-02-05 14:42:28

:o_____________________nice

irspow
2006-02-05 02:38:52

Quadratic equations are of the form:

ax² + bx + c = y,   a, b, and c are just the constants before the variables.

Like a would be the constant term before the x² and so forth.

Your equation above does not describe something falling however.

1) There is no initial height.

2) The initial velocity would suggest that it was going up and not down (though you can have positive values for problems like this)

3) If we are using gravity, then the acceleration would be negative and not positive.

A free-fall position formula should take the form:

Yf = Yi + vt - .5gt²;  Yf = final height, Yi = initial height, v = initial velocity

The way that you have it written states that:

Yi = 40m,  v = +10m/s,  a = -9.8m/s²

So,

0 = 40 + 10t - 4.9t² (when it hits the ground Yf = 0)

-4.9t² + 10t + 40 = 0

[-10 ± √(100 + 784)] / -9.8

t = -2.01348... and 4.054299...

Since time cannot be negative;

t ≈ 4.0543s

Sorry Mathsyperson, you forgot to use the correct conditions stipulated by the problem.  Don't let Mathsy's error confuse you about his math ability, he is very astute.

2006-02-05 02:10:48

thanks a lot but what is a b and c?

mathsyperson
2006-02-05 01:30:28

Just plug everything into the equation you've got.

s=40
u=10
a=9.8 (acceleration due to gravity close to the Earth's surface)
t=?

40 = 10t + 0.5*9.8*t²

4.9t² + 10t - 40 = 0

That's just a quadratic equation, which you can easily solve using the quadratic formula:

Substituting the values gives an answer of 2.01 seconds, to 3 significant figures.

2006-02-05 00:48:10

a body moving with uniform acceleration exhibits the relationship

s=ut+0.5at^2

s= distance
u= initial velocity
a= acceleration
t= time

if an object falls and passes a point 40m above ground at a velocity of 10m/s how long does it take to hit the ground by: