Take Sally's position to be the origin, and up-the-ramp to be the positive direction. The ball travels a distance <em>x</em> in time <em>t</em> of
<em>x</em> = <em>u</em> <em>t</em> + 1/2 (- 3.7 m/s²) <em>t</em>²
where <em>u</em> is the ball's initial velocity.
Its velocity <em>v</em> at time <em>t</em> is
<em>v</em> = <em>u</em> + (- 3.7 m/s²) <em>t</em>
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Let <em>T</em> be the time it takes for the ball to reach the second person 19.6 m up the ramp. At this time, the ball attains a velocity of 4.9 m/s, so that
4.9 m/s = <em>u</em> + (- 3.7 m/s²) <em>T</em>
<em>T</em> = (<em>u</em> - 4.9 m/s) / (3.7 m/s²)
Substitute this into the distance equation, with <em>x</em> = 19.6 m, and solve for <em>u</em> :
19.6 m = <em>u</em> (<em>u</em> - 4.9 m/s) / (3.7 m/s²) + 1/2 (- 3.7 m/s²) ((<em>u</em> - 4.9 m/s) / (3.7 m/s²))²
<em>u</em> ≈ 13 m/s
