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Mathematics 28 Online
OpenStudy (anonymous):

A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise? (b) how long does it take to reach its highest point?(c) hOW LONG DOES THE BALL TAKE TO HIT THE GROUND AFTER IT REACHES ITS HIGHEST POIN?(d) what is its velocity when it return to the level from it started?

OpenStudy (anonymous):

whr r u stuck

OpenStudy (anonymous):

don't know where to start basic physic class

OpenStudy (anonymous):

u know first equation of motion?

OpenStudy (anonymous):

second n third too lol

OpenStudy (anonymous):

Vf^1?

OpenStudy (anonymous):

what??

OpenStudy (anonymous):

no

OpenStudy (anonymous):

An object moving with a constant accelerationa (sometimes called a uniform acceleration) starts with an initial velocity u and achieves a final velocity v in a time of t seconds. The quantities u, v, a, and t are linked mathematically by the equation: v= u + at This is called the first equation of motion.

OpenStudy (anonymous):

My professor didn't teach us that I just know its a free falling question and I'm clueless

OpenStudy (anonymous):

okk, u use v^2 = u^2 - 2gs

OpenStudy (anonymous):

would it be 2(25.0)

OpenStudy (anonymous):

no v = 0 u = 25 g = 10 find s

OpenStudy (anonymous):

would it be 31.25

OpenStudy (anonymous):

awsumm correct

OpenStudy (anonymous):

would that be the answer to part (a)?

OpenStudy (anonymous):

yepp, for b part use the first equation of motion: v = u - gt

OpenStudy (anonymous):

Wow you explain this a lot better where it makes sense than my professor Thank you so much :D

OpenStudy (anonymous):

what formula would I use for (c)?

OpenStudy (anonymous):

you re welcome first u tell me what all vaues u know for c part

OpenStudy (anonymous):

Okay writing problem down and equation real quick sorry

OpenStudy (anonymous):

Is t=2.5 answer for part (b)?

OpenStudy (anonymous):

awsumm, correct

OpenStudy (anonymous):

so part c would I simply multiply 31.25 x2.5?

OpenStudy (anonymous):

no, u never do taht in free fall second equation of motion this time: s = ut + 1/2g t^2 notice the plus sign coz we r goin down

OpenStudy (anonymous):

would it look like this 31.25=25(2.5)+.5(10)(2.5)^2

OpenStudy (anonymous):

yes, but t is to be found it is not given while falling down interestingly, it ll come out to be 2.5 lol

OpenStudy (anonymous):

okay I had 62.5/31.5 is that right?

OpenStudy (anonymous):

I meant 31.25

OpenStudy (anonymous):

no sqrt 62.5/10

OpenStudy (anonymous):

and thnx for the testimonial

OpenStudy (anonymous):

You Should be a professor

OpenStudy (anonymous):

hahaa, i am not a professor, but i do teach physics in my institute n my classes r very hip lol

OpenStudy (anonymous):

What would formula set up be for part D and I promise that's it ;D

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