Projectile Motion with Basketball

In summary, a ball released from a height of 1.3 m above the ground with a horizontal speed of 1.45 m/s will have a horizontal distance of 0.36 m and a vertical height of 0.99 m after 0.25 seconds. After 0.5 seconds, the ball will have traveled a horizontal distance of 0.73 m and a vertical height of 0.24 m. At 0.5 seconds, the ball's total velocity will be 4.8 m/s with a direction of 73.5 degrees below the horizontal.
  • #1
wolves5
52
0
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.
 
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  • #2
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
 

1. What is projectile motion with basketball?

Projectile motion with basketball is the curved path that a basketball takes when it is thrown or shot through the air. It is a combination of horizontal and vertical motion, influenced by gravity.

2. How is projectile motion with basketball different from other sports?

Unlike other sports, the trajectory of a basketball is affected by the spin and bounce of the ball, as well as the force and angle of the shot. Other sports, such as baseball or football, have more predictable trajectories due to the shape and size of the ball.

3. What factors affect the trajectory of a basketball?

The trajectory of a basketball is affected by the initial velocity, angle of projection, air resistance, and gravity. The shape and texture of the ball can also play a role in the trajectory.

4. How does air resistance impact the flight of a basketball?

Air resistance, also known as drag, can slow down the basketball and change its trajectory. This is why basketballs with different textures or patterns may have slightly different trajectories when thrown with the same force and angle.

5. Can you predict the exact trajectory of a basketball?

It is difficult to predict the exact trajectory of a basketball due to the many variables that can affect its flight. However, with knowledge of the initial conditions and a good understanding of the principles of projectile motion, we can make fairly accurate predictions of the general path of the basketball.

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