A 4kg ball is on a 5 m ledge. If it is pushed off the ledge, how much kinetic energy will it have just before hitting the ground?

196 J

Explanation:

From the law of conservation of energy,

ΔU + ΔK = 0 where ΔU = change in potential energy = mgΔy and ΔK = change in kinetic energy.

ΔK = -ΔU

If we take the top of the ledge as y₁ = 5 and the ground as y₂ = 0, then Δy = y₂ - y₁ = 0 - 5 = -5 m

ΔK = -ΔU = -mgΔy = -mg -5 = 5mg = 5 m × 4 kg × 9.8 m/s = 196 J

ΔK = K₂ - K₁ where K₁ = initial kinetic of 4 kg ball = 0 J (since it is initially at rest on the ledge) and K₂ = final kinetic energy of 4 kg ball before it hits the ground.

ΔK = K₂ - K₁ = 196 J  

K₂ - 0 = 196 J

K₂ = 196 J

So, it has a kinetic energy of 196 J before it hits the ground

Explanation:

Given:

Mass, M = 4 kg

Height, S = 5 m

Initial velocity, u = 0 m/s (at rest)

Using equation of motion,

V^2 = u^2 + 2a × S

= 0 + 2 × 9.8 × 5

= 98

V = sqrt(98)

= 9.889 m/s

= 9.9 m/s

Kinetic energy, E = 1/2 × mass × v2

Note: initial kinetic energy = 0 because velocity = 0 m/s(at rest)

Final kinetic energy, Ef = 1/2 × 4 × 9.9^2

= 196 J

= 0.196 kJ

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

find the sum of the two horizontal vectors and the sum of the two vertical vectors

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