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From the edge of a cliff, a 0.49 kg projectile is launched with an initial kinetic energy of 1670 J. The projectile's maximum upward displacement from the launch point is 160 m. What are the (a) horizontal and (b) vertical components of its launch velocity?

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princessesther2011

Answer:

(a) Horizontal component of the launch velocity = 60.7m/s

(b) Vertical component of the launch velocity = 56.0m/s

Explanation:

Initial Kinetic Energy = (IKE)= 1670J, mass (m) = 0.49kg, hmax = 160m

IKE = 1/2mu^2

1670×2/0.48 = u^2

u^2 = 6816.3

u = √6816.3 = 82.6m/s

hmax = u^2sin^2A/2g

160×2×9.8/6816.3 = sin^2A

Sin^2A = 0.46

SinA = √0.46 = 0.6782

A = inverse (sin 0.6782)

A = 42.7° (angle of inclination to the horizontal)

(a) Horizontal component (Ux) = ucosA = 82.6×cos42.7° = 82.6×0.7349 = 60.7m/s

(b) Vertical component (Uy) = usinA = 82.6×sin42.7° = 82.6×0.6782 = 56.0m/s

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