When an object falls from a h height, you should work
with the uniformly accelerated linear movement equations:
y=½*a*t²+Vo*t+yo
You should consider:
a=-g=-10m/s²
yo=h
If it’s a freefall, it means it starts from rest, which
means it has no initial velocity:
Vo=0
Replacing that information in the equation:
y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h
So this is the
Besides, if you want to find out how long it takes for it
to get to the floor, you should put the height of the floor as final height,
which would be 0 (assuming the initial height has been measured from there):
y=0
0=-5m/s²*t²+h
5m/s²*t²=h
t²=h/(5m/s²)
t=√(h/(5m/s²))
t=√(hs²/(5m))
t=(√(h/(5m)))s
<span>If
we <span>quadruple
</span>h:</span>
t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1
This 4 goes inside the square root, so then it converts
to 2. So the new time is twice as much the previous time.
Concerning velocity, you have to use the other equation:
v=at+vo
As I said before, a is gravity and vo is zero.
v=-10m/s²*t+0=-10m/s²*t
Final velocity is directly related to time, so if time is
doubled, so is velocity.
v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1
<span>So the correct answer is A, and the other ones are
false.</span>
