Velocity of the bowling ball- projectile motion

AI Thread Summary
Tony drops a bowling ball from a height of 0.28m, and it travels a horizontal distance of 1.80m before hitting the floor. The initial calculations suggest a horizontal velocity of 0.43m/s, but further analysis indicates a velocity of 7.53m/s when considering the time taken for the ball to fall. The final velocity calculation, which combines horizontal and vertical components, results in 7.89m/s, but this includes gravitational effects, which may not be relevant to the original question. The main concern is to determine the initial horizontal velocity at which Tony threw the ball, which should be 7.53m/s, excluding gravitational acceleration. The discussion emphasizes the importance of distinguishing between the effects of gravity and the initial throw velocity.
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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!
 
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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!
 
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