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...
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.
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.
thanks
Sure! Let's start with the given equation:
X = Xo + Vo t + (1/2) a t^2
To rearrange this equation for t, we need to isolate t on one side of the equation. Here's how we can do that step by step:
Step 1: Move all terms without t to the other side of the equation:
X - Xo = Vo t + (1/2) a t^2
Step 2: Factor out t from the terms on the right-hand side:
X - Xo = t (Vo + (1/2) a t)
Step 3: Divide both sides of the equation by the expression in front of t to solve for t:
(X - Xo) / (Vo + (1/2) a t) = t
Step 4: Simplify the right-hand side by distributing:
(X - Xo) / (Vo + (1/2) a t) = t
(X - Xo) = t(Vo + (1/2) a t)
Step 5: Distribute t on the right-hand side:
(X - Xo) = Vo t + (1/2) a t^2
Now, we have a quadratic equation. To solve for t, we can rearrange it as follows:
Step 6: Move all terms to one side of the equation:
0 = (1/2) a t^2 + Vo t + (Xo - X)
Step 7: Multiply through by 2 to eliminate the fraction:
0 = a t^2 + 2 Vo t + 2(Xo - X)
Now, this equation is in the form of ax^2 + bx + c = 0, where a = a, b = 2 Vo, and c = 2(Xo - X). We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / 2a
Using the values of a, b, and c from Step 7, we can plug them into the quadratic formula to find the values of t.
To find the velocity just before "landing" (or when t = 0), you can substitute t = 0 into the original equation:
X = Xo + Vo t + (1/2) a t^2
Since the term with t^2 becomes 0, the equation simplifies to:
X = Xo + Vo(0)
Therefore, the velocity just before "landing" is Vo.
I hope this step-by-step explanation helps you understand how to rearrange the equation for t and solve the problem!