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
free lil peezy and drake is kool
Assuming this is an elastic collision, they go in the direction of the object with more mass
<span>In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a C. series of compressions and rarefactions.</span>
No, the building's size in comparison to the earth would have no change or change so increadibly miniscule, like if you were told to spin slowly and an and was placed on top of your head