Question

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

sense Xo = 0 m
sense Vo = 0 s

I can write

X = 2^-1 a t^2

solve for t

(X = 2^-1 a t^2)2
(2x = a t^2)a^-1
(a^-1 2x = t^2)^(2^-1)
(a^-1 2x)^(2^-1) = t

plug and chug

((9.80 s^-2 m)^-1 2(380 m))^(2^-1)
t is about 7.93 s

the unit check gives seconds

my text book provides the answer of 8.81 s in the apendex and my teacher said that the answer is just under 9 s I do not understand

Answer 1

oh... Xo is not zero

Answer 2

X = Xo + Vo t + 2^-1 a t^2

sense Vo = 0 s^-1 m
sense x = 0 m

0 = Xo + 2^-1 a t^2

0 - Xo = Xo - Xo + 2^-1 a t^2
(- Xo = 2^-1 a t^2)a^-1 2
a^-1 (2-Xo) = t^2)^(2^-1)
(a^-1 (2-Xo))^(2^-1) = t

that would give me a negetive number =_=

Answer 3

imaginary number =_+

Answer 4

ok i accept that this is true

0 = Xo + 2^-1 a t^2

and then I solved for t
0 - Xo = Xo - Xo + 2^-1 a t^2
(- Xo = 2^-1 a t^2)a^-1 2
a^-1 (2-Xo) = t^2)^(2^-1)
(a^-1 (2-Xo))^(2^-1) = t

and i do not see what i did wrong

Answer 5

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.
Refer to the calculations of
http://www.jiskha.com/display.cgi?id=1245771533

Answer 6

t= 6.3
v=60.76 m/s