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A person walking with a speed of 1.45 m/sec releases a ball from a height of h = 1.3 m above the ground. Use the point on the ground, directly below where the ball is initially released at the origin of your coordinate system.

Picture: https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1201/fall-evening/homework/Ch-03-04/wt-bouncing-ball/bouncing-ball.jpg [Broken]

(a) What is the ball's position at 0.25 seconds after it is released in the x and y direction?

x =

y =

(b) What is the ball's position at 0.5 seconds after it is released in the x and y direction?

x =

y =

(c) What is the ball's total velocity, speed and direction of motion at 0.5 seconds after it is released?

vx =

vy =

v =

Θ =

For a, I tried using vf= vi + at. I used gravity (9.81) for the acceleration. However, its not right. For part b, I understand that you would use the same equation as part A, but you would just switch the 0.25 seconds to 0.5 seconds. For part c, I understand you use sin and cosine in order to solve for vx and vy.

Picture: https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1201/fall-evening/homework/Ch-03-04/wt-bouncing-ball/bouncing-ball.jpg [Broken]

(a) What is the ball's position at 0.25 seconds after it is released in the x and y direction?

x =

y =

(b) What is the ball's position at 0.5 seconds after it is released in the x and y direction?

x =

y =

(c) What is the ball's total velocity, speed and direction of motion at 0.5 seconds after it is released?

vx =

vy =

v =

Θ =

For a, I tried using vf= vi + at. I used gravity (9.81) for the acceleration. However, its not right. For part b, I understand that you would use the same equation as part A, but you would just switch the 0.25 seconds to 0.5 seconds. For part c, I understand you use sin and cosine in order to solve for vx and vy.

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