A 5-gram [g] rubber ball is released from a height of 8 meters [m] above a flat surface on the moon. Gravitational accelerati
on on the moon is 1.62 meters per second squared [m/s2]. Assume that no energy is lost from frictional drag. What is the velocity, in units of meters per second [m/s], of the rubber ball the instant before it strikes the flat surface?
2 answers:
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Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s