(a) Just for fun, a person jumps from rest from the top of a cliff overlooking a lake. In falling through a distance H, she acquires a certain speed v. Assuming free-fall conditions, how much farther must she fall in order to acquire speed 2v? Express your answer in terms of H.
I found its 3H
Well, if we assume free-fall conditions, I guess we can also assume the person is falling headfirst, right? That's an important detail, you know. Safety first! Now, let's do some math.
When the person falls through a distance H and acquires the speed v, we can use the equation of motion:
v² = 2gh
Where g is the acceleration due to gravity. Now, we want to find out how much farther the person needs to fall to acquire a speed of 2v. Let's call that distance X.
So, when the person falls through a distance H + X, the speed acquired will be 2v. Using the same equation as before:
(2v)² = 2g(H + X)
Now, let's compare those two equations:
v² = 2gh
(2v)² = 2g(H + X)
Dividing the second equation by the first equation:
(2v)² / v² = (2g(H + X)) / (2gh)
4 = (H + X) / H
Cross-multiplying:
4H = H + X
Now, let's solve for X:
4H - H = X
3H = X
So, the person must fall an additional 3H in order to acquire a speed of 2v. Voila!
But remember, gravity is a serious business, and I wouldn't recommend jumping off cliffs just for fun. Leave the jumping to the kangaroos!
To find out how much farther the person must fall in order to acquire a speed of 2v, we need to understand the relationship between the distance fallen and the speed acquired during free fall.
When an object falls freely under the influence of gravity, it experiences constant acceleration. This acceleration is approximately equal to the acceleration due to gravity, denoted as 'g' which is about 9.8 meters per second squared (m/s^2) near the Earth's surface.
Let's use the following equations of motion for free fall:
1. Velocity (v) after falling a distance (h) can be calculated using the equation:
v^2 = u^2 + 2gh
where,
v = final velocity
u = initial velocity (which is 0 because the person is at rest)
g = acceleration due to gravity
h = distance fallen (which is denoted as H in the question)
Since the person already acquires a speed of v after falling through a distance H, we can write the equation as:
v^2 = 0 + 2gH
v^2 = 2gH ----(equation 1)
Now, we want to find out how much farther the person must fall in order to acquire a speed of 2v. Let's denote this additional distance as 'x'.
After falling through a total distance of H + x, the speed acquired will be 2v. Using the same equation, we can write:
(2v)^2 = 0 + 2g(H + x)
4v^2 = 2g(H + x)
2v^2 = g(H + x) ----(equation 2)
To find x, we can rearrange equation 2 as follows:
2v^2 - gH = gx
x = (2v^2 - gH) / g
= 2v^2/g - H
Therefore, the person must fall an additional distance of 2v^2/g - H in order to acquire a speed of 2v.
Soko
Four times farther, since the kinetic energy must quadruple.
4 H