16. The speed of an object falling from rest is directly proportional to the square root of the distance the object has fallen. When an object has fallen 36 ft , its speed is \( 48 \mathrm{ft} / \mathrm{s} \). How much farther must it fall before its speed is \( 80 \mathrm{ft} / \mathrm{s} \) ?

To find out how much farther the object must fall to reach a speed of \( 80 \, \text{ft/s} \), we can use the relationship given: speed \( v \) is directly proportional to the square root of the distance \( d \) fallen. The relationship can be expressed as: \[ v = k \sqrt{d} \] where \( k \) is the proportionality constant. From the problem statement, we know that when \( d = 36 \, \text{ft} \), \( v = 48 \, \text{ft/s} \). Plugging in these values, we can solve for \( k \): \[ 48 = k \sqrt{36} \] \[ 48 = k \cdot 6 \implies k = \frac{48}{6} = 8 \] Now we have the equation: \[ v = 8 \sqrt{d} \] Next, we want to find \( d \) when \( v = 80 \, \text{ft/s} \): \[ 80 = 8 \sqrt{d} \] \[ \sqrt{d} = \frac{80}{8} = 10 \implies d = 10^2 = 100 \, \text{ft} \] This represents the total distance fallen to reach \( 80 \, \text{ft/s} \). We already know that the object has fallen \( 36 \, \text{ft} \), so to find the additional distance it needs to fall, we subtract the distance already fallen from the total distance: \[ \text{Additional distance} = 100 \, \text{ft} - 36 \, \text{ft} = 64 \, \text{ft} \] Thus, the object must fall an additional \( 64 \, \text{ft} \) to reach a speed of \( 80 \, \text{ft/s} \).