Answer:
11.3 m/s
Explanation:
KE₁ = KE₂
½m₁v₁² = ½m₂v₂²
½ (2 kg) v² = ½ (4 kg) (8 m/s)²
v ≈ 11.3 m/s
Explanation:
Given parameters:
Distance hopped = 84m
Displacement = 84m due east
Time = 7s
Unknown:
Speed of kangaroo = ?
Velocity of kangaroo = ?
Solution:
To solve this problem,
Speed =
=
= 12m/s
Velocity =
=
= 12m/s due east
"Constant velocity" is practically a definition for zero acceleration.
Answer:
39.7 m
Explanation:
First, we conside only the last second of fall of the body. We can apply the following suvat equation:

where, taking downward as positive direction:
s = 23 m is the displacement of the body
t = 1 s is the time interval considered
is the acceleration
u is the velocity of the body at the beginning of that second
Solving for u, we find:

Now we can call this velocity that we found v,
v = 18 m/s
And we can now consider the first part of the fall, where we can apply the following suvat equation:

where
v = 18 m/s
u = 0 (the body falls from rest)
s' is the displacement of the body before the last second
Solving for s',

Therefore, the total heigth of the building is the sum of s and s':
h = s + s' = 23 m + 16.7 m = 39.7 m