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A hopper jumps straight up to a height of 1.1 m. With what velocity did it leave the floor?
1 answer:
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Answer:
= 2,018.3 J
Explanation:
mass (m) = 1.1 kg
velocity (v) = 62 m/s
initial angle(θ) = 12.3°
kinetic energy =
where
- m is the mass
- v is the velocity, which has two components
v = Vx + Vy (horizontal component + vertical component)
at the maximum height, velocity is 0, so Vy = 0 and v= Vx
v = Vx = v.cosθ
v = 62 x cos 12.3 = 60.6 m/s
therefore kinetic energy = =
= 2,018.3 J