Basketball game problem- projectile motion

Click For Summary
SUMMARY

The basketball projectile motion problem involves calculating the initial speed and launch angle required for a basketball to reach the hoop exactly one second before the game ends. The player throws the ball from a height of 206 cm towards a basket 306 cm high, located 26 m away. The calculations yield an initial horizontal velocity (v0x) of 13 m/s and a vertical velocity (v0y) of 5.4 m/s, resulting in an initial velocity (v0) of 14.08 m/s and a launch angle (α) of 22.56 degrees. The discussion emphasizes the need to consider potential energy changes when determining final velocity and impact angle.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with kinematic equations
  • Knowledge of energy conservation principles
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Explore advanced projectile motion simulations using tools like PhET Interactive Simulations
  • Learn about energy conservation in projectile motion scenarios
  • Investigate the effects of air resistance on projectile trajectories
  • Study the mathematical derivation of projectile motion equations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of projectile motion in sports contexts, particularly in basketball.

mmoadi
Messages
149
Reaction score
0

Homework Statement



Basketball player throws the ball form the middle of the court one second before the game ends. How much must the initial speed of the ball be and at what angle it must be thrown, if it should hit the basket at the exact time of siren without preceding reflection from board? Length of playground is 26 m, the distance between the floor and the ring is 306 cm, and the ball is thrown from height 206 cm. Neglect the rotation of the ball. At what angle and with what velocity (speed) the ball hits the basket?

Homework Equations



Δx= v0xt
Δy= v0yt – ½(gt²)
v0= sqrt(v0x² + v0y²)
tan α= v0y / v0x

The Attempt at a Solution



What I know:
t= 1 s
L= 26 m
Δx= 13 m
h1= 206 cm
h2= 306 cm
Δy= 100 cm = 1 m

What I’m looking for:
v0= ? m/s (initial velocity)
α = ? ° (launch angle)
vf= ? m/s (final velocity)
β= ? ° (impact angle)

① First, I calculated the v0x = vx: v0x= Δx / t = 13 m/s
② Second, I calculated v0y: v0y= (2Δy + gt²) / 2t= 5.4 m/s
③ I know calculated v0 from the Pythagoras formula: v0= sqrt(v0x² + v0y²)= 14.08 m/s
④ And know I calculated the launching angle α: α= (tan)-¹(v0y / v0x)= 22.56°

What is left know are final velocity and impact angle. Can I just state that final velocity = initial velocity (conservation of energy) and that impact angle = launch angle (symmetry of parabola)? Or, do I have to do new calculations because the basketball ring is higher than the launching height of the ball and the parabola is not symmetrical? If second is the case, I really need help, because I don’t know how to approach it now.

Thank you for helping!
 
Last edited:
Physics news on Phys.org
You can't say that final velocity = initial velocity (conservation of energy), because
you have to account for the increase in potential energy of the ball.

another option is to use V_f = V_i + a t and compute a new v_x and v_y
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
36K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
4K
  • · Replies 18 ·
Replies
18
Views
5K