A 11-kg cannonball is fired with a velocity of 19m / s from 28 m above the ground . It hits a target that is 11 m above the ground . With what velocity does it hit, in m / s

Question

A 11-kg cannonball is fired with a velocity of 19m / s from 28 m above the ground . It hits a target that is 11 m above the ground . With what velocity does it hit, in m / s ? (Round your answer to one decimal place if necessary .)

Answer 1

I assume we are to ignore air resistance.

In that case we can use conservation of energy:

The cannon ball's energy at the beginning is the same as at the end.

At both points, the energy is the sum of potential and kinetic energy.

(For potential energy we need a reference point of 0 potential energy,

and I will use the ground for that.)

Let

m = 11 kg be the mass of the cannon ball,

v₀ = 19 m/s be the initial velocity,

h₀ = 28 m be the initial height,

v₁ (to be calculated) be the final velocity,

h₁ = 11 m be the final height,

g = 9.81 m/s² be gravitational acceleration,

P₀ = mgh₀ be the initial potential energy,

K₀ = mv₀²/2 be the initial kinetic energy,

P₁ = mgh₁ be the final potential energy,

K₁ = mv₁²/2 be the final kinetic energy.

By conservation of energy

P₀ + K₀ = P₁ + K₁

Substitute

mgh₀ + mv₀²/2 = mgh₁ + mv₁²/2 (1)

From that express

v₁ = √(2g(h₀ - h₁) + v₀²)

Substitute actual numbers

v₁ = √(2×9.81×(28 - 11) + 19²) ≈ 26.4 m/s

Notice that we did not need to know the mass of the cannon ball.

The result would be the same for cannon balls of any size because

we can divide equation (1) by m.