Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Here you go mate. Hope it helps u. Pls follow me in reddit lol username: RobloxNoob2006
To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.
By definition we know that
Where,
KE =Kinetic Energy
KR = Rotational Kinetic Energy
PE = Potential Energy
In this way
Where,
m = mass
v= Velocity
I = Moment of Inertia
Angular velocity
g = Gravity
h = Height
We know as well that 
for velocity (v) and Radius (r)
Therefore replacing we have
![[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2](https://tex.z-dn.net/?f=%5Btex%5Dh%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7BI%7D%7Bmg%7D%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2)
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Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s
Answer:
firstly,
ke=1÷2mv^2
on putting same ke by increasing mass by 16 times new velocity becomes v'
then
ke=1÷2×16m×v'^2
from above we can write
1÷2mv^2=1÷2×16m×v'^2
v^2=16v'^2
1÷4v=v'
hence original velocity should be decreased by 4 times to keep same ke
