# Word problems on projectile motion and geometry

1. A ball thrown vertically upward is caught by the thrower after 4.00 s.

(a) Find the initial velocity of the ball.

m/s

(b) Find the maximum height it reaches.

m

2. A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash. The first stone had an initial velocity of -2.40 m/s.

(a) How long after release of the first stone did the two stones hit the water?

s

(b) What initial velocity must the second stone have had, given that they hit the water simultaneously?

m/s

(c) What was the velocity of each stone at the instant it hit the water?

first stone m/s

second stone m/s

3. A person walks 25.0° north of east for 3.00 km. How far would the person walk due north and due east to arrive at the same location?

north km

east km

4. A small map shows New York to be 486 miles in a direction 8° north of east from Columbus. The same map shows that Buffalo is 302 miles in a direction 56° west of north from New York. Assume a flat earth and use the given information to find the displacement from Columbus to Buffalo.

magnitude miles

direction ° north of east of Columbus

5. An airplane starting from airport A flies 300 km east, then 355 km at 24.0° west of north, and then 150 km north to arrive finally at airport B.

(a) The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight?

° north of east

(b) How far will the pilot travel in the flight? Assume there is no wind during either flight.

km

https://brainmass.com/physics/velocity/word-problems-on-projectile-motion-and-geometry-95119

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1. A ball thrown vertically upward is caught by the thrower after 4.00 s.

(a) Find the initial velocity of the ball.

m/s

(b) Find the maximum height it reaches.

m

By symmetry, it takes 4/2 = 2s for the ball to reach it's maximum height and then to fall from max. height to the thrower. Also at the max. height its velocity = 0.

Use v = u + at upward.

0 = u - 9.8*2

u = 19.6 m/s

Use v2 = u2 + 2as upward.

0 = 19.62 - 2*9.8*s

s = 19.6 m

2. A mountain climber stands at the top of a 40.0 m cliff that ...

#### Solution Summary

I have provided detailed solution to five word problems. Solution is in a 4-page word document.