So you fall from 1500m and you gain 9.8m/s/s...how long til you reach the ground?

Question

I figured it to be 17 seconds, but how do I put that in equation form and go through it?

d=sxt

1500=9.8m/s/s x t

1500/9.8m/s/s = t

but the 9.8 is not constant, so how does it work.

Thanks!

Answer 1

s = ut + 1/2 at²

where s = displacement (here 1500 m )

u = initial velocity (here 0 m/s)

a = acceleration (here 9.8 m/s²)

t = time (to find)

So,

1500 = 0 + 1/2 * 9.8 * t²

=> 3000/9.8 = t²

=> t² = 306.122

=> t = 17.49 seconds

Answer 2

The first answer is correct.  But I need to comment on your question because its wording indicates some uncertainty about the physics.

When you say "you gain" 9.8 m/sec^2, you are indicating an increase in the acceleration rate.  In fact, the average acceleration due to gravity for a free falling object in a vacuum is constant, not gaining, at g = 9.81 m/sec^2 on Earth's surface.  Because a = g = constant = 9.81 m/sec^2 we can do the following.

Average velocity (V - U)/2 = V/2; where U = 0 the initial velocity at S = 1,500 meters and V = ? the terminal velocity before going splat on the cement.  Without drag, the terminal velocity V = gt where t is the time to fall and g is the constant acceleration due to gravity; so V/2 = (1/2)gt.  That is, the average velocity in a dragless free fall from zero initial velocity is just 1/2 the terminal velocity.

The distance traveled during the fall is S = (V/2)t, which is the average velocity falling for the t time.  Combining all this, we have S = (V/2)t = (1/2)gt^2, which is what the first answer indicated.  You can solve for t.

But now you know why...now you know the physics for the answer.  It all stems from the average velocity while falling a time t.  And the average velocity is possible because the average g = 9.81 m/sec^2 is fixed on Earth's surface.

Answer 3

fall=1500 meters = 1/2 * g * t^2

t = sqr(3 000 / 9.8) = 17.50 seconds

g=9.8 meters/second^2

g is a constant "close to the earth's surface"

Answer 4

Divide 15000 by 4.9 and find the square root, it will be 17.5. This is the time, in seconds, an object will be in free fall from this altitude, disregarding air Resistance.

Answer 5

d = V0(t) + 1/2a(t²)

d = distance

V0 = initial velocity (assuming 0 in this problem)

t = time until you die (I mean reach the ground)

a = acceleration (due to gravity, g = 9.8 m/s²)

1500 = 0 + 1/2(9.8)t²

1500 = 4.9t²

t² = 306.12

t = +/- √306.12

t = 17.5 s or -17.5s

Since scientist have concluded that time travel is impossibe, t can't be negative, so t = 17.5s

If you want the formula, you will need to solve for t:

d = 1/2a(t²)

2d = at²

t² = 2d/a

t = √(2d/a)

This formula only works if V0 is 0. Otherwise, you would have to use the quadratic equation:

d = V0(t) + 1/2at²

1/2at² + V0(t) - d = 0

t = (-V0 +/- √[V0² - 4 (1/2a)(-d)])/(2*1/2a)

t = -V0/a +/- [√(V0² + 2ad)]/a