Question

I GIVE MEDALS.

A gargoyle statue falls off the side of a building from a height of 75 m. (Assume no air resistance.)
How long does it take the statue to reach the ground?

I think i use the equation d=1/2(a)(t2)+v1(t)
but im not sure where to plug in all the numbers to solve.

Answer 1

You have the right formula, you just need to plug in
d = 75 m
a = 9.8 ms^-2
v1 = 0

Answer 2

how would the equation look rearranged so that it's t=? @broken_symmetry

Answer 3

Well let's start with
\[d = \frac12 a t^2\]
Multiply both sides by 2/a
\[\frac{2d}a = t^2\]
take the square root of both sides
\[\sqrt\frac{2d}a = t\]

Answer 4

so the final answer would be 3.9? @broken_symmetry

Answer 5

Yeah
http://www.wolframalpha.com/input/?i=75+%3D+4.9*x^2

Answer 6

how would i figure out how fast is the statue moving when it reaches the ground? @broken_symmetry

Answer 7

Well, you know the acceleration (9.8 ms^-2) and you just worked out the time. You know it was stationary to begin with.
So...

Answer 8

is it the same equation? @broken_symmetry

Answer 9

No, that equation was to get distance.
The equation for velocity from acceleration and time is much simpler.

Answer 10

is it just displacement divided by time? @broken_symmetry

Answer 11

That gets you the average velocity, but you want the final velocity.

Acceleration is velocity divided by time.

There's a clue in the units
\[\text{m s}^{-2}\]

So if something has been accelerating at 9.8 ms^-2 for 3.9 s, how fast is it now going?

Answer 12

it would just be 38.22 correct? @broken_symmetry

Answer 13

Yep :)

Answer 14

Just a word of advice: I found it really really useful to memorise the SUVAT equations:
\[v = u+at\]
\[s=ut+\frac12at^2 = vt-\frac12at^2\]
\[v^2=u^2+2as\]
\[s=\frac{u+v}{t}\]
s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

You can choose which one to use by which one you know all the variables except the one you want to work out.

Answer 15

thanks so much! @broken_symmetry