The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to
<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²
and
<em>v</em> = <em>v</em>₀ - <em>g t</em>
where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.
The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that
-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²
-7.94 m/s = <em>v</em>₀ - <em>g t</em>
Solve for <em>t</em> in the second equation:
<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>
Substitute this into the first equation and solve for <em>v</em>₀ :
-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²
-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>
2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²
2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)
-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²
<em>v</em>₀² = 9.73 m²/s²
<em>v</em>₀ = 3.12 m/s
is given by
the specific heat capacity
and the amount of heat supplied is 
, so if we re-arrange the previous formula we find the increase in temperature of the water: