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.(adsbygoogle = window.adsbygoogle || []).push({});

Picture: https://wug-s.physics.uiuc.edu/cgi/...k/Ch-03-04/wt-bouncing-ball/bouncing-ball.jpg

(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.

**Physics Forums - The Fusion of Science and Community**

# Projectile Motion with Basketball

Have something to add?

- Similar discussions for: Projectile Motion with Basketball

Loading...

**Physics Forums - The Fusion of Science and Community**