Projectile Motion with Basketball


by wolves5
Tags: basketball, motion, projectile
wolves5
wolves5 is offline
#1
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.

Picture: https://wug-s.physics.uiuc.edu/cgi/c...ncing-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.
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mburt
mburt is offline
#2
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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