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

Click For Summary
To determine the speed needed for a basketball to reach a hoop 2.8 m away and 0.4 m high, the player throws the ball at a 60° angle. The equations of motion, including V=V0+at and r(t)=r0+v0t+0.5a^2, were applied, but an error in dimensional analysis led to an incorrect solution. Correcting the equation to properly reflect the relationship between height and velocity was crucial. General advice emphasized working with symbols before substituting numerical values to avoid mistakes. Following this guidance resulted in finding the correct answer.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
3
Views
4K
  • · Replies 18 ·
Replies
18
Views
4K
  • · 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