# 4 Problems: Vectors, Displacement and equations of motion.

Please see the attached file for full problem description.

1.) A basketball player runs down the court, following the path indicated by the vectors A, B, and C in Figure 3-38. The magnitudes of these three vectors are: A = 10.0 m, B = 21.0 m, and C = 7.0 m. Let the +x axis point to the right and the +y axis point to the far side of the court.

a.) Find the magnitude and direction of the net displacement of the player using the component method of vector addition.

2.) Two of the allowed chess moves for a knight are shown below

a.) Is the magnitude of displacement 1 greater than, less than, or equal to the magnitude of displacement 2?

b.) Find the magnitude and direction of the knight's displacement for each of the two moves. Assume that the checkerboard squares are 3.39 cm on a side.

Magnitude 1 ________cm

Direction 1 ________ degrees (counterclockwise from the x-axis)

Magnitude 2 ________cm

Direction 2 ________ degrees (counterclockwise from the x-axis)

3.) At the 18th green of the U. S. Open you need to make a 20.0 ft putt to win the tournament. When you hit the ball, giving it an initial speed of 1.67 m/s, it stops 6.00 ft short of the hole.

a.) Assuming the deceleration caused by the grass is constant, what should the initial speed have been to make the putt?

_____m/s

b.)What initial speed do you need to just make the remaining 6.00 ft putt?

4.) Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the floor 2.5 s after first striking it, what was the ball's greatest height above the floor?

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#### Solution Preview

Please see the attached file.

1.) A basketball player runs down the court, following the path indicated by the vectors A, B, and C in Figure 3-38. The magnitudes of these three vectors are: A = 10.0 m, B = 21.0 m, and C = 7.0 m. Let the +x axis point to the right and the +y axis point to the far side of the court.

Figure 3-38

a.) Find the magnitude and direction of the net displacement of the player using the component method of vector addition.

Corrected

A(x)=0 B(x) = 21 cos (45) = 14.85 m C(x) =7 cos (30) = 6.06 m

A(y) = - 10 m B(y) = 21 sin(45) = 14.85 m C(y)= -7sin(30) = -3.5 m

So then

D(x) = 0 + 14.85 + 6.06 = 20.91 m

D(y) = - 10 + 14.85 - 3.5 = 1.35 m

Then:

D= square root [(20.91)^2 + (1.35)^2] = 20.95 m

Theta = arc tan (1.35/20.91)= 3.69 degrees

2.) Two of the allowed chess moves for ...

#### Solution Summary

4 solution good to understand displacement, velocity acceleration with equations of motion and motion under gravity,