Hello, UMStudent!

1) Two balls are thrown upward from the edge of the cliff.

The first is thrown with a speed of 48 ft/sec

and the second one is thrown a second later with a speed of 24 ft/sec.

Do the balls ever pass each other?

Let the top of the cliff be y = 0.

At time t, the first ball has height: .y .= .48t - 16t²

. . and the second ball has height: .y .= .24(t - 1) - 16(t - 1)²

They will pass when they both have the same height.

. . 48t - 16t² .= .24(t - 1) - 16(t - 1)²

This simplifies to: .8t = 40 . → . t = 5

The second ball overtakes the first ball 5 second after the first ball is thrown,

. . at a height 160 feet below the top of the cliff.

2) A woman at a point A on the shore of a circular lake with radius 2 mi wants to

arrive at the diametrically opposite point C on the other side of the lake in the

shortest possible time. .She can walk at the rate of 4 mph and row a boat at 2 mph.

How should she proceed? Code:

B
* * *
* \ *
* \ *
* \ *
\
* θ \ *
A * - - - - * - - - - * C
* 2 O 2 *
* *
* *
* *
* * *

Draw radius OB = 2.

Let __/__AOB = θ

She will walk along the circular arc AB.

The length of AB is 2θ miles (where θ is in radians).

At 4 mph, it will take her: .2θ/4 = ½θ hours to walk the arc.

She will row across the chord BC.

The length of the chord can be found with the Law of Cosines.

. . BC² .= .2² + 2² - 2(2)(2)cos(2π - θ) .= .8 + 8·cos θ .= .8(1 + cos θ)

. . BC² .= .16[(1 + cos θ)/2] .= .16·cos²(½θ)

. . Hence: .BC .= .4·cos(½θ) miles.

At 2 mph, it will take her: .4·cos(½θ)/2 = 2·cos(½θ) hours to row across.

Her total time is: .T .= .½θ + 2·cos(½θ) hours

And **that** is the function we must minimize . . .