What Speed Is Needed for a Basketball to Reach the Hoop?

Click For Summary
SUMMARY

The discussion centers on calculating the required speed for a basketball to reach a hoop located 2.8 meters away and 0.4 meters high, thrown at a 60° angle. The correct approach involves using the equations of motion, specifically V=V0+at and r(t)=r0+v0t+0.5a^2, while ensuring proper dimensional analysis. A common mistake identified was the incorrect substitution of terms, specifically using (5.6 V_0)^2 instead of the correct ({5.6/V_0})^2. Following the advice to work with symbols before substituting values led to the correct answer.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions, particularly sine
  • Ability to perform dimensional analysis
NEXT STEPS
  • Study projectile motion in physics, focusing on angle and distance calculations
  • Learn about dimensional analysis techniques to avoid common errors
  • Explore the use of kinematic equations in different scenarios
  • Practice solving problems involving angles and heights in projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for effective problem-solving strategies in kinematics.

agargento
Messages
38
Reaction score
2

Homework Statement


A basketball player throws the ball at a 60° angle above the horizontal to a hoop which is located a horizontal distance L = 2.8 m from the point of release and at a height h = 0.4 m above it. What is the required speed if the basketball is to reach the hoop?

HlInqYZ.gif


Homework Equations


I used V=V0+at and r(t)=r0+v0t+0.5a2
and V0 x/y=V0sin(θ)

The Attempt at a Solution


My attempt: http://i.imgur.com/0vdqdvl.jpg
0vdqdvl.jpg

Got a wrong answer. Would love some guidance. Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
**Third question** You write ##0.4 = ... (5.6 V_0 )^2 ## instead of ##0.4 = ... ({5.6 \over V_0}) ^2 ##

A quick check on the dimensions of the last term would have shown that it's not length like the other two.

General advice is to work with symbols and only substitute values at the very last.
 
BvU said:
**Third question** You write ##0.4 = ... (5.6 V_0 )^2 ## instead of ##0.4 = ... ({5.6 \over V_0}) ^2 ##

A quick check on the dimensions of the last term would have shown that it's not length like the other two.

General advice is to work with symbols and only substitute values at the very last.

Thank you so much! Got the right answer, and definitely will use your advice in the future.
 

Similar threads

Replies
3
Views
4K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 6 ·
Replies
6
Views
10K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K