4). A coyote on jet skies chases a roadrunner. The Coyote\'s inital velocity is

Question

4). A coyote on jet skies chases a roadrunner. The Coyote's inital velocity is a zero and his initial and costant acceleration is 15 m/s^2. If the coyote is 120 m from the same direction, (a) what would the roadrunner's constant speed have to be if he is to reach the edge of the cliff (where he reverses his direction) at the same time as the coyote? (b) If the cliff is 100m above the floor of the cayon, how far would he coyote land from the base of the cliff? (c) What are the components of the coyote's velocity just before he hits the ground?

5) A plane flies from base camp to lake. A. 288 km in a direction 22 degrees N of E. Tje plane flies to lake B, which is 120 Km from lake A and in a direction 34 degrees W of N. Use the law of cosines or vector addition to determine the planes' final position with respect to its starting position.

6) If a pitcher throws a baseball at 101 mi/h, how far would the ball fall vertically after traveling 66m horizontally?

Explanation / Answer

6 ) Ignoring significant figures (which we shouldn't do, but whatever), and assuming mph means miles per hour

we convert 101 miles per hour to 45.25 m/s. So when the ball has traveled 66 m horizontally, about 1.45 seconds have passed.

The only vertical motion is due to gravity, which provides constant acceleration of approximately -9.8 m/s^2. We can use the equation

s = si + vi (delta t) + 1/2 * a * (delta t)^2

Rewrite this as

s - si =vi (delta t) + 1/2 * a * (delta t)^2

Since s - si will be the change from initial position to the position at this time.

As an approximation, we can assume that vi = 0 (in the vertical direction, since nearly all initial velocity is forward). So vi = 0. Then s - si = 1/2 * (-9.8 m/s^2) * (1.45)^2

s - si = -10.3

This means (si - s) = 10.3. The ball has dropped approximately 10.3m.

5 )

First find the i & j components of the individual vectors.
(Camp -> Lake A)
i = 288 cos 22 = 267
j =288 sin 22 = 107.88
(Lake A -> Lake B)
i = 120 cos 120 = -60
j = 120 sin 120 = 103.92

Add the i's and j's together.
i = 267 - 130 = 137
j = 107.8 + 103.92 = 211.72

Distance = 252.1km
Note: If you had converted everything to m before any calculations, you would have ended up with 252.1km.
Direction = tan-1 (211.72 / 137) = 57.09degrees.

4 )

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