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Projectile Motion with Basketball 
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#1
Jun111, 06:07 PM

P: 52

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://wugs.physics.uiuc.edu/cgi/c...ncingball.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. 


#2
Jun111, 06:19 PM

P: 52

a) Finding the horizontal distance after 0.25 s is fairly straightforward:
Given: v_{x} = 1.45 m/s d_{x} = v_{x}t = (1.45 m/s)(0.25 s) = 0.36 m Finding the vertical height at 0.25 s: Given: v_{1y} = 0 d_{y} = 0 + 0.5a(t)^2 = 0.5(9.8)(0.25)^2 = 0.31 m 1.3 m  0.31 m = 0.99 m b) ... same process c) d_{y} = 0 + 0.5at^2 = 0.5(9.8)(0.5)^2 = 1.23 m v_{2y}^{2}  v_{1y}^{2} = 2a_{y}d_{y} v_{2y}^{2} = 2(9.8)(1.23) + (0) v_{2y} = 4.9 m/s v_{2x} = 1.45 m/s (horizontal speed is constant throught) Then do a Pythagorean triangle: v^2 = (4.9)^2 + (1.45)^2 v = 4.8 m/s tanx = 4.9/1.45 x = 73.5 degrees below the horizontal 


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