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

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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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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 a knight are shown below

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

The first displacement is 2 units (blocks) right and one unit toward us thus the magnitude or resultant will be

The second displacement is 1 unit (blocks) left and two units toward us thus the magnitude or resultant will be

Hence magnitude of both displacements is same.

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: 7.58 cm

As above the magnitude will be

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

The angle will be arc tan (1/2) = 26.57 degree
Hence direction with positive x axis will be - 26.57 deg or 360 - 26.57 = 333.43 degrees
Magnitude 2: 7.58 cm

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

The angle will be arc tan (2/1) = 63.43 degree
Hence direction with positive x axis will be 180 + 63.43 = 243.43 degrees

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

Initial velocity u = 1.67 m/s
Final velocity v = 0
Displacement s = 20 - 6 = 14 ft = 14*0.3048 = 4.27 m

Using third equation of motion the acceleration is given by

Gives a = - 0.327 m/s2
For the second case
Initial velocity u = ? m/s
Final velocity v = 0
Displacement s = 20 ft = 20*0.3048 = 6.01 m

Using third equation of motion again the initial velocity is given by

Gives u = 2.0 m/s

b.)What initial speed do you need to just make the remaining 6.00 ft putt?
Initial velocity u =? m/s
Final velocity v = 0
Displacement s = 6 ft = 6*0.3048 = 1.83 m

Using third equation of motion again the initial velocity is given by

Gives u = 1.09 m/s

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?
Neglecting air resistance the time taken by the ball to move up will be same as coming down and hence the time of fall from the maximum height will be 2.5/2 = 1.25 s
Considering downward positive
Initial velocity u = 0 m/s
Displacement h =?
Acceleration a = g = 9.8 m/s2
Time t = 1.25 s

Using second equation of motion the maximum height reached is given by

Hence the maximum height reached will be 7.66 m

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 5, 2022, 4:37 pm ad1c9bdddf>
https://brainmass.com/physics/velocity/problems-vectors-displacement-equations-motion-200813