g A projectile of mass 3 kg is launched horizontally from an initial height 3 m with an initial velocity 10 m/s. This velocity i
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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
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s