Finding Initial Velocity with no time or other Velocities

Click For Summary

Homework Help Overview

The problem involves a basketball player attempting to determine the initial speed required for a basketball to reach a basket located horizontally 5.52 m away and 3.05 m above the floor. The ball is released from a height of 1.52 m at an angle of 32.0 degrees above the horizontal. The challenge lies in calculating the initial velocity without having the time or other velocities provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the horizontal and vertical components of motion to find time and initial velocity. There are attempts to express time in terms of initial velocity and to relate the components of motion through equations. Some participants express uncertainty about the correctness of their derived equations and the values obtained.

Discussion Status

The discussion is ongoing with participants exploring different equations and relationships between the variables. Some have provided guidance on how to approach the problem, while others are questioning their calculations and the assumptions made in their reasoning. There is no explicit consensus on the correct initial speed yet.

Contextual Notes

Participants are working under the constraints of not having time or other velocities directly given, which complicates their calculations. There are indications of confusion regarding the application of trigonometric functions and the interpretation of the equations derived.

Hypnos_16
Messages
148
Reaction score
1

Homework Statement


An Olympic basketball player shoots towards a basket that is 5.52 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.52 m above the floor at an angle of 32.0o above the horizontal. What initial speed should she give the ball so that it reaches the basket and hopefully scores?

all i can get out of this is
dx = 5.52m
dy = 3.05 though she's only shooting 1.53m to the basket
Theta = 32 degs
ay = -9.81

Homework Equations


I don't know how to get the time out of this problem, nor the velocity
any time i try I'm told it would be easier to merge two equations, though i don't know how.
Please someone help.


The Attempt at a Solution

 
Physics news on Phys.org
Welcome to PF!

Hi Hypnos_16! Welcome to PF! :smile:

(have a theta: θ :wink:)

Call the speed v, and use the x components to find t.

Then use that value of t in the y components to find how high the ball is.

Show us what you get. :smile:​
 
so I've tried writing out the formula. i got this
x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1

but i don't know what to do from there, i don't know even if that's right, and if it is where do i plug that into?
 
Hypnos_16 said:
so I've tried writing out the formula. i got this
x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1

(try using the X2 icon just above the Reply box :wink:)

good so far, now your v1 in that equation has to be the initial component of velocity in the x direction, which is v times … ? :smile:
 
times v by cos of 32 degs! then you get t with a value v that you can use to find v1! okay i think i get it now

t = 5.52/v1cos32
t = 6.51v1

x = x1 + v1t
5.51 = 0 + v1 x 6.51v1
5.51 = 6.51v1^2
v1 = 0.92m/s

i got this, but it doesn't seem right.
i then got a time of 6 seconds, which seems to high.
 
Last edited:
Hypnos_16 said:
times v by cos of 32 degs!

That's it! :wink:

If you have any difficulty with the rest of the problem, just come back and tell us. :smile:
 
so i got the v in the x direction as 0.92m/s the time in the air to be 6 seconds but the v in the y direction seems outrageously high. I think i goofed up somewhere.

x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1
t = 5.52/v1cos32
t = 6.51v1

x = x1 + v1t
5.51 = 0 + v1 x 6.51v1
5.51 = 6.51v1^2
v1 = 0.92m/s

t = 6.51(0.92)
t = 6s

dy = y1 + v1t -gt^2
3.05 = 1.53 + sin32v1(6) - 4.9(36)
1.52 = sin32v1(6) - 176.4
177.92 = sin32v1(6)
177.92 = 3.18v1
55.9 = v1
 
Last edited:
Hypnos_16 said:
x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1
t = 5.52/v1cos32
t = 6.51v1

No, the last two lines should be
t = 5.52/vcos32
t = 6.51/v, shouldn't it?

(where v is the total initial speed)

Now use that value of t (ie use 6.51/v ) in an equation for the y direction. :smile:
 
so after i fill that in for t do i have to work out the result as a quadratic equation?
dy = vy1 + v1t - gt^2
3.05 = 1.53 + sin32(6.51v) - 4.9(6.51v)^2
1.52 = 3.45v - 207.7v^2
0 = 1.52 + 3.45v - 207.7v^2
that's the answer i got.
would finding this result in finding the initial overall speed or are there more steps?
 
  • #10
(try using the X2 icon just above the Reply box :wink:)
Hypnos_16 said:
dy = vy1 + v1t - gt^2
3.05 = 1.53 + sin32(6.51v) - 4.9(6.51v)^2
1.52 = 3.45v - 207.7v^2
0 = 1.52 + 3.45v - 207.7v^2

No, it's 6.51/v, not 6.51 times v. :redface:

(and i think you got your 1.53 and 1.52 the wrong way round)
would finding this result in finding the initial overall speed or are there more steps?

we defined v as the initial overall speed, didn't we? :rolleyes:
 
  • #11
so how would you multiply sin32(6.51/v) and 4.9(6.51/v)^2?
 
  • #12
Hypnos_16 said:
so how would you multiply sin32(6.51/v) and 4.9(6.51/v)^2?

uhh? :confused:

it's just (sin32*6.51)/v and (4.9*(6.51)2)/v2.

get some sleep! :zzz:​
 
  • #13
will do, after i get this one done
so after all that work i got 11.7m/s
does that seem like it would be the right answer?
 
  • #14
Hypnos_16 said:
will do, after i get this one done
so after all that work i got 11.7m/s
does that seem like it would be the right answer?

erm :redface: … I don't mind checking your calculations, but I'm not going to do the work myself! :wink:
 
  • #15
No that's fine you don't have to do it for me, i just want to make sure I've done it correctly, make sure the math is right.
 
  • #16
Hypnos_16 said:
No that's fine you don't have to do it for me, i just want to make sure I've done it correctly, make sure the math is right.

Well, the general technique is right, but you keep making mistakes like v instead of 1/v, so it's difficult to say without seeing the whole thing.

I'm off to bed now, anyway …

goodnight! :zzz:​
 

Similar threads

Calculating Initial Velocity for a Basketball Free Throw - Projectile Motion
  • · Replies 3 ·
Replies
3
Views
4K
Initial velocity given time and angle
  • · Replies 8 ·
Replies
8
Views
2K
Calculating Initial Speed for an Olympic Basketball Shot
  • · Replies 6 ·
Replies
6
Views
2K
Find Initial Velocity of Projectile Basketball
  • · Replies 6 ·
Replies
6
Views
10K
Calculating Initial Velocity of a Basketball Shot
  • · Replies 1 ·
Replies
1
Views
2K
Projectile motion - finding initial velocity with no time
  • · Replies 3 ·
Replies
3
Views
10K
Solving Basketball Parabola Puzzle: Initial Velocity at 35°
  • · Replies 1 ·
Replies
1
Views
4K
Throw a ball, initial velocity?
  • · Replies 11 ·
Replies
11
Views
5K
Solving an Initial Velocity Problem for a Basketball Player
  • · Replies 2 ·
Replies
2
Views
7K
Finding initial velocity of a cat jumping through a hoop
  • · Replies 1 ·
Replies
1
Views
2K