# Problem of the Day: 1/18/13

The height of a falling object to Earth (barring air resistance) is $h=-16t^2+i$, where i is the initial height and t is the time in seconds after the object is dropped. After how many seconds will an object land on the ground if it is dropped from a plane 3,600 feet in the air?

Solution to yesterday’s problem:

The volume of a cone is $V=\frac{1}{3}\pi r^3 h$. So the volume of the original cone is

$V_1=\frac{1}{3}\pi (2)^3 (10)$

$V_1=\frac{1}{3}\pi (80)$

$V_1=\frac{80\pi}{3}$

The top 5 cm of the cone is cut off. Since the height is half the original height, the radius is half the original radius. The volume of the smaller cone is

$V_2=\frac{1}{3}\pi (1)^3 (5)$

$V_2=\frac{1}{3}\pi (5)$

$V_2=\frac{5\pi}{3}$

Subtracting the smaller cone’s volume from that of the larger cone, we get

$V_3=\frac{80\pi}{3}-\frac{5\pi}{3}$

$V_3=\frac{75\pi}{3}$

$V_3=25\pi$

So the volume of the frustum is $25\pi$, or approximately 78.54 cubic cm.