Projectile motion when you are only given time and range

Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem involving a golf ball launched at an angle of 30 degrees, with a horizontal range of 150 meters. Participants are tasked with finding the initial velocity, time of flight, and maximum height of the projectile.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the initial velocity using a formula related to range and angle. There are attempts to determine the x and y components of the initial velocity and to identify the relevant kinematic equations for time of flight and maximum height.

Discussion Status

Some participants have provided calculations and confirmations regarding the initial velocity and components of motion. There is ongoing exploration of how to derive time of flight and maximum height, with some guidance offered on using horizontal displacement and velocity components.

Contextual Notes

Participants express uncertainty about the formulas for time of flight and maximum height, indicating a reliance on notes that do not contain this information. There is also mention of deriving equations based on knowledge of vectors and kinematics.

Acmnji
Messages
11
Reaction score
0

Homework Statement


I am very lost and would love help. Given in the description is;"A golf ball leaves a golf club at an angel 30 degrees above the horizontal and lands on the green 150 meters away initial position." With this information i am supposed to find the balls initial velocity, how long it was in the air and how high it rises

The Attempt at a Solution


to find the initial velocity i did delta r = [vi^2 * Sin (60) ]/ 9.81 = 41.22.
not 100% sure this is correct but its what i have so far and i have no idea where to go from here
 
Physics news on Phys.org
Acmnji said:

Homework Statement


I am very lost and would love help. Given in the description is;"A golf ball leaves a golf club at an angel 30 degrees above the horizontal and lands on the green 150 meters away initial position." With this information i am supposed to find the balls initial velocity, how long it was in the air and how high it rises

The Attempt at a Solution


to find the initial velocity i did delta r = [vi^2 * Sin (60) ]/ 9.81 = 41.22.
not 100% sure this is correct but its what i have so far and i have no idea where to go from here
Your calculation of initial velocity is correct.

You need to find the time of flight (T) and maximum height (H). Do your notes contain the formulas for T and H ?
 
cnh1995 said:
Your calculation of initial velocity is correct.

You need to find the time of flight (T) and maximum height (H). Do your notes contain the formulas for T and H ?

okay thanks and they don't contain either of those formulas and i don't see either in my notes as well.
 
Acmnji said:
okay thanks and they don't contain either of those formulas and i don't see either in my notes as well.
Well, that's odd!

Anyway, you can look them up or you can derive them on your own if you applied your knowledge of vectors and kinematic equations.
What are the x and y components of initial velocity? What is the acceleration in x and y directions?
 
for the x and y components of vi would it be 41.22cos30 and 41.22sin30?
 
Acmnji said:
for the x and y components of vi would it be 41.22cos30 and 41.22sin30?
Yes.
What about acceleration?
 
no acceleration in the x and a acceleration of -9.81 in the y?
 
Acmnji said:
no acceleration in the x and a acceleration of -9.81 in the y?
Correct.
Can you write the relevant kinematic equations for finding T and H?
 
im sorry but i have no clue and I am not able to find it anywhere.
 
  • #10
Acmnji said:
im sorry but i have no clue and I am not able to find it anywhere.
What is the horizontal dispalcement? Which component of velocity is responsible for it?
 
  • #11
vix and time dictate horizontal displacement
 
  • #12
Acmnji said:
vix and time dictate horizontal displacement
Yes.
So you can find time from this. You know Vx and horizontal displacement.
 
  • #13
So if the horizontal displacement is 150 and the vx is 35.7 it would be 150/35.7=time=4.2 seconds?
 
  • #14
Acmnji said:
So if the horizontal displacement is 150 and the vx is 35.7 it would be 150/35.7=time=4.2 seconds?
Yes.
 
  • #15
and the equation for height, would that be vy + (0.5*9.81*4.2^2) ?
 
  • #16
Acmnji said:
and the equation for height, would that be vy + (0.5*9.81*4.2^2) ?
Careful there..
At what time will the ball be at the maximum height?
 
  • #17
at half the time so 2.1 seconds?
 
  • #18
Acmnji said:
at half the time so 2.1 seconds?
Right.
 
  • #19
im confused now, how would that change the equation
 
  • #20
Acmnji said:
vy + (0.5*9.81*4.2^2) ?
You have taken t=4.2s..
 
  • #21
so then would it change to vy + (0.5*9.81*2.1^2)
 
  • #22
Ok. Would you have access to a graphing calculator compare the path of the golf ball to your answer?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
997
Replies
40
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
8
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
27K