During an energetic game of catch in the Wren courtyard, a ball crashes through President Sullivan's second story window (height = 6 m above the ground), with a speed when it hits the window of 12 m/s. It was thrown from a height 2 m above the ground. At what speed was it thrown?

Solution:

While this is an example of projectile motion, we are not give enough information to apply the usual projectile motion equations (we don't know the mass of the ball, the angle it was thrown at, nor the horizontal distance travelled). So, apply energy conservation instead (after all, it was an energetic game of catch...):

E = (1/2)mv₁² + mgh₁ = (1/2)mv₂² + mgh₂

I have chosen U=0 at the ground, so h₁ = 2 m and h₂ = 6 m (one could just as easily choose the zero of potential to be the orginal height of the ball, and we would have h₁ = 0 m and h₂ = 4 m). We are given v₂ = 12 m/s. The mass m cancels everywhere in the equation, so the only unknown is v₁. Rearranging the energy equation we have

v₁² = v₂² + 2gh₂ - 2gh₁

v₁ = [v₂² + 2g(h₂ - h₁)]^1/2

= [(12 m/s)² + 2 (9.8 m/s²)(6 m - 2 m)]^1/2

= 14.9 m/s

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