Velocity of the bowling ball- projectile motion

Click For Summary
SUMMARY

The discussion centers on calculating the initial horizontal velocity of a bowling ball dropped by Tony from a height of 0.28 meters, which travels a horizontal distance of 1.80 meters before hitting the ground. The correct calculation yields a horizontal velocity of 7.53 m/s, derived from the time taken to fall and the horizontal distance covered. Additionally, the final velocity of the ball, calculated as 7.89 m/s, includes vertical motion due to gravity, which is not relevant to the original question. The consensus is that the initial velocity should only account for the horizontal motion.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of horizontal and vertical velocity components
  • Ability to apply significant figures in calculations
NEXT STEPS
  • Study kinematic equations for projectile motion in detail
  • Learn how to separate horizontal and vertical motion in physics problems
  • Explore the concept of initial velocity in projectile motion
  • Review significant figures and their importance in scientific calculations
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of objects in motion.

dani123
Messages
134
Reaction score
0

Homework Statement



Tony has a tendency to drop his bowling ball on the floor at the completion of his follow-through swing. Tony let's the ball go when the ball is 0.28m above the floor. If the ball travels a horizontal distance of 1.80m before you hear a "thud", with what velocity did Tony throw the bowling ball?

Homework Equations



dv=1/2*at2
dh=Vh*Δt
Kinetic equation d=Vi*t+ 1/2*at2
dh=-V2*sin2θ /g

The Attempt at a Solution



dv=0.28m
dh=1.80m
Vh=?

Solve for t and found that t=4.18s by rearranging the dv equation.

Then that value was plugged into the dh equation and rearranged to solve for Vh, to give Vh=0.43m/s.

I would like for someone to just verify my answer and check that the significant figures are being respected. Thanks so much in advance!
 
Physics news on Phys.org
Is the initial velocity of the ball horizontal?
 
it doesn't specify in the question.. but I that's what I think... :S
 
Right, since that the max speed of the ball will be when the ball hits the floor and travels horizontal, you'll need to find the time that this takes. You have an a=0, u=0 S(vertical)=0.28m.

From this you can use s=ut+1/2at2 Since U=0 the time that this takes is equal to √2s/a.

You now have a time, and a horizontal distance, use speed =horizontal distance/time to get horizontal velocity.

This works out to be 7.53 meters per second. Hope that helps.
 
I did the same thing when i used an example i found online as a guideline, but then i went one step further and found the final velocity of the entire motion i think. After finding 7.53m/s, i went on to do the following:

|Vf|= √(Vix)2+(Vfy)2= √(7.53m/s)2+(-2.34m/s)2= 7.89m/s

but now I am confused as to why I did that last part and if even means anything... Do you know what this last part I did gives me? If it even gives me any information in regards to Tony and the way he throws the bowling ball?

Thank you very much for your time and help, it is greatly appreciated!
 
That finds the magnitude of it, though I think this would include the motion due to Earth's gravity. Which isn't what the question wants. it wants how fast did he throw the ball. If anything you need to discount the velocity that the ball would gain due to gravity. Though I'm fairly confident in 7.53.
 
thank you!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
4
Views
3K
Replies
19
Views
3K