If
X = vertical distance from Xo at time t, positive upwards, and
Vo = velocity at t=0
a = acceleration due to gravity, -9.81 m/s/s
Then the proposed equation is correct.
X = Xo + Vo t + 2^-1 a t^2
For your particular case,
Xo=380 m, (top of building)
X=0 (ground)
Vo=0 (just before the fall)
a=-9.81 m/s/s
That translates to
0 = 380 + 0*t + (-9.81)t2/2
Since the middle term "disappears" because of the coefficient of 0, then
9.81t2/2 = 380
from which you can find t without much ado.
The time is a touch more than 6 seconds if air resistance and his exceptional height are ignored.
I'm trying to solve this problem
Calculate (a) how long it took King Kong to fall straight down form the top of the Empire State Building (380 m high), and (b) his velocity just before "landing".
My teacher said
try and rearagne this equation for t
X = Xo + Vo t + 2^-1 a t^2
I tried to rearange it for t and i guess i did it wrong i got
t = a^-1 (2(X - Xo - Vo))
I guess i did it wrong
so could you please show me how to rearange the equation for t step by step because I guess I did it wrong thanks...
3 answers
thanks
Sorry, t should be a touch less than 9 seconds, being
sqrt(2*380/9.81)
The 6 seconds estimate did not account for the factor of 2.
sqrt(2*380/9.81)
The 6 seconds estimate did not account for the factor of 2.