Projectile Motion with Basketball

by wolves5
Tags: basketball, motion, projectile
wolves5 is offline
Jun1-11, 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.


(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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mburt is offline
Jun1-11, 06:19 PM
P: 52
a) Finding the horizontal distance after 0.25 s is fairly straightforward:

Given: vx = 1.45 m/s

dx = vxt
= (1.45 m/s)(0.25 s) = 0.36 m

Finding the vertical height at 0.25 s:

Given: v1y = 0

dy = 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) dy = 0 + 0.5at^2 = 0.5(-9.8)(0.5)^2
= -1.23 m

v2y2 - v1y2 = 2aydy
v2y2 = 2(-9.8)(-1.23) + (0)
v2y = 4.9 m/s
v2x = 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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