After what time interval does the ball strike the ground

AI Thread Summary
A ball is thrown downward from a height of 30m with an initial speed of 8m/s, and the acceleration due to gravity is -9.8m/s². To determine the time it takes for the ball to strike the ground, the appropriate equation is Yf = Yi + Vi(t) + 1/2g(t²), while ensuring the correct signs for direction are used. The discussion also briefly touches on a separate problem involving a horse moving at 10m/s, but confusion arises regarding the context of the "limb" referenced. The recommended approach for the horse's problem is to use a distance formula, as time is not provided but distance is. Overall, clarity in problem statements is emphasized to avoid confusion in calculations.
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Homework Statement


A ball is thrown downward with initial speed of 8m/s from a height of 30m.

Homework Equations


\After what time interval does the ball strike the ground



The Attempt at a Solution



acceleration will equal -9.8m/s^2 because it's free fall
would i use equation V=Vo-gt










constant speed of a horse is 10m/s. distance from limb to saddle is 3m

A) find horizontal distance between saddle and limb?
B) for what time interval is he in the air?

Attempted solution:
X=Xo+Vo(t)+V2(a)(t^2)
what equation do you use to find the time ... Yf=Yi+Vi(t)+1/2g(t^2) and rearange it to
 
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I'm confused. There seems to be two questions... one about a ball... about about a horse and saddle?
 
sorry. it's two different problems
 
012435 said:
sorry. it's two different problems

This is the right equation to use for the first problem:

Yf=Yi+Vi(t)+1/2g(t^2)

but you need to be careful about directions and signs... you've taken downwards as positive here...
 
012435 said:

Homework Statement


A ball is thrown downward with initial speed of 8m/s from a height of 30m.

Homework Equations


\After what time interval does the ball strike the ground



The Attempt at a Solution



acceleration will equal -9.8m/s^2 because it's free fall
would i use equation V=Vo-gt
Since you are NOT given the time but ARE given a distance, it would be better to use a distance formula: d= (1/2)gt2+ v0t. Donht forget that v0 is negative.








constant speed of a horse is 10m/s. distance from limb to saddle is 3m

A) find horizontal distance between saddle and limb?
B) for what time interval is he in the air?

Attempted solution:
X=Xo+Vo(t)+V2(a)(t^2)
what equation do you use to find the time ... Yf=Yi+Vi(t)+1/2g(t^2) and rearange it to
This makes no sense to me at all. What does the "distance from saddle to limb" . At some paces some horses can go 10m/s while always having at least one foot on the ground- they are never "in the air".

And what "limb"? A foreleg? Or is this a limb of a tree? Did you leave something out? is the horse jumping over a limb?
 
HallsofIvy said:
Since you are NOT given the time but ARE given a distance, it would be better to use a distance formula: d= (1/2)gt2+ v0t. Donht forget that v0 is negative.

yea i would use this formula
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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