Question

The figure shows the trajectory of a golf ball in projectile motion over level ground. The time t0 = 0 s corresponds to the moment just after the ball is launched from position x0 = 0 m, y0 = 0 m. Its initial velocity (launch velocity) is v0. Two other points along the trajectory are indicated in the figure. At time t1 (with velocity v1), the ball reaches the peak of its trajectory. Its position at this moment is denoted by (x1, y1) = (x1, ymax), since it is at its maximum height. At time t2 (with velocity v2), just before the ball touches the ground again, its position is (x2, y2) = (xmax, y2), since it is at its maximum horizontal range. 9. [1pt] Which of the statements are true? E.g., enter AB. A. 0 > v0,y = v1,y = v2,y B. a1,y = 0 > a0,y = a2,y = -g C. a0,y = a1,y = a2,y = -g D. |v0| = |v2| > |v1| > 0 E. v0,x > v1,x = 0 > v2,x F. |v0| > |v1| > |v2| > 0 G. v0,y > v1,y = 0 > v2,y H. v0,x = v1,x = v2,x > 0 Correct, computer gets: CDGH 10. [1pt] Assume an initial speed of 22.2 m/s, at an angle of 37.1° with respect to the positive x axis. Find the velocity at time t0; enter the x component, then the y component. Answer 1 of 2: Answer 2 of 2: Submit All Answers

          The figure shows the trajectory of a golf ball in projectile motion over level ground. The time t0 = 0 s corresponds to the moment just after the ball is launched from position x0 = 0 m, y0 = 0 m. Its initial velocity (launch velocity) is v0. Two other points along the trajectory are indicated in the figure. At time t1 (with velocity v1), the ball reaches the peak of its trajectory. Its position at this moment is denoted by (x1, y1) = (x1, ymax), since it is at its maximum height. At time t2 (with velocity v2), just before the ball touches the ground again, its position is (x2, y2) = (xmax, y2), since it is at its maximum horizontal range.

9. [1pt]
Which of the statements are true? E.g., enter AB.

A. 0 > v0,y = v1,y = v2,y
B. a1,y = 0 > a0,y = a2,y = -g
C. a0,y = a1,y = a2,y = -g
D. |v0| = |v2| > |v1| > 0
E. v0,x > v1,x = 0 > v2,x
F. |v0| > |v1| > |v2| > 0
G. v0,y > v1,y = 0 > v2,y
H. v0,x = v1,x = v2,x > 0

Correct, computer gets: CDGH

10. [1pt]
Assume an initial speed of 22.2 m/s, at an angle of 37.1&deg; with respect to the positive x axis. Find the velocity at time t0; enter the x component, then the y component.

Answer 1 of 2:
Answer 2 of 2:
Submit All Answers

The figure shows the trajectory of a golf ball in projectile motion over level ground. The time t0 = 0 s corresponds to the moment just after the ball is launched from position x0 = 0 m, y0 = 0 m. Its initial velocity (launch velocity) is v0. Two other points along the trajectory are indicated in the figure. At time t1 (with velocity v1), the ball reaches the peak of its trajectory. Its position at this moment is denoted by (x1, y1) = (x1, ymax), since it is at its maximum height. At time t2 (with velocity v2), just before the ball touches the ground again, its position is (x2, y2) = (xmax, y2), since it is at its maximum horizontal range.

9. [1pt]
Which of the statements are true? E.g., enter AB.

A. 0 > v0,y = v1,y = v2,y
B. a1,y = 0 > a0,y = a2,y = -g
C. a0,y = a1,y = a2,y = -g
D. |v0| = |v2| > |v1| > 0
E. v0,x > v1,x = 0 > v2,x
F. |v0| > |v1| > |v2| > 0
G. v0,y > v1,y = 0 > v2,y
H. v0,x = v1,x = v2,x > 0

Correct, computer gets: CDGH

10. [1pt]
Assume an initial speed of 22.2 m/s, at an angle of 37.1 deg; with respect to the positive x axis. Find the velocity at time t0; enter the x component, then the y component.

Answer 1 of 2:
Answer 2 of 2:
Submit All Answers

Added by Brian G.

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