Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ba
2 answers:
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Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.