Question

question: you drop a 14 - kg ball from a height of 1.5 m, and it only bounces back to a height of 0.85 m. what was the magnitude of the impulse on the ball when it hit the floor?

Explanation:

Step1: Find velocity before impact

Use kinematic equation $v^2 = u^2 + 2gh$. Initial velocity $u=0$, $g=9.8\ \text{m/s}^2$, $h_1=1.5\ \text{m}$.
$v_1 = \sqrt{2gh_1} = \sqrt{2\times9.8\times1.5} = \sqrt{29.4} \approx 5.422\ \text{m/s}$ (downward, negative sign: $v_1 \approx -5.422\ \text{m/s}$)

Step2: Find velocity after impact

Use same kinematic equation for rebound, final velocity at peak $u=0$, $h_2=0.85\ \text{m}$.
$v_2 = \sqrt{2gh_2} = \sqrt{2\times9.8\times0.85} = \sqrt{16.66} \approx 4.082\ \text{m/s}$ (upward, positive sign)

Step3: Calculate change in momentum

Impulse $J = \Delta p = m(v_2 - v_1)$, $m=14\ \text{kg}$.
$J = 14\times(4.082 - (-5.422)) = 14\times(9.504)$

Step4: Compute final impulse value

$14\times9.504 = 133.056$

Answer:

$\approx 133\ \text{kg·m/s}$ (or $133.1\ \text{kg·m/s}$ for more precision)