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
6 answers
oh... Xo is not zero
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 =_=
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 =_=
imaginary number =_+
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
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
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
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
t= 6.3
v=60.76 m/s
v=60.76 m/s
6 answers