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## catsrule332 3 years ago A ball is thrown straight up at 15m/s. How long does the ball take to reach apex? What is the ball's total time of flight?

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1. cinar

$v^2=v_0^2-2gh$ v=0 v_0=15 find h then use this formula to find out t $h=v _{i}t-\frac12g t^2$

2. catsrule332

I found the total time it takes the ball to reach the apex but can't find the total time in flight?

3. cinar

it is 2 times reach to apex

4. catsrule332

awesome! thanks

5. cinar

there is something wrong

6. cinar

what did you find t

7. catsrule332

1.5 s and total time 3 s

8. cinar

how about h?

9. cinar

is it 11.5

10. cinar

because t is not real, it is complex..

11. cinar

its initial speed is 15 right, can you check it again?

12. cinar

actually there is a formula which says how long take to reach apex it is t=v_i/g t=15/9.8=1.5 sec and there for you seem to right..

13. cinar

but I'am still curies why that eq. did not work

14. JamesJ

Yes, you actually don't need to find the height which is reaches. You have that v(t) = u - gt where v(t) is the velocity in the vertical direction at time t and g is gravitational accel

15. JamesJ

hence it's at the apex when v(t) = 0.

16. cinar

what if I want to solve problem using by those formulas, it must be worked, where am I making wrong?

17. JamesJ

Using g = 10 m/s^2, the max height is given by 2gh = v^2 h = v^2/2g = 15^2/20 = 11.25 Now there is a solution to the position equation h = vt - gt^2/2 i.e., 11.25 = 15t - 10t^2 That solution is t = 1.5

18. cinar

no it is complex, and when I put t=1.5 right side is 0 left side is 11.25 and also $\Delta <0$

19. JamesJ

Right. I should have written not 10t^2, but 5t^2, because the coefficient of t^2 is g/2, not g. Hence the equation is $11.25 = 15t - 5t^2$ which is equivalent to $5t^2 - 15t + 11.25 = 0$ and that quadratic has discriminant of $\Delta = (-15)^2 - 4(5)(11.25) = 0$

20. cinar

that's interesting when I accept g=9.8 it is not real

21. cinar
22. cinar

maybe their teacher said them to take g=10

23. chuchay

what is the formula use in pascal's law?

24. JamesJ

If you use any value of g > 0, we from the Conservation of Energy that $v_i^2 = 2gh$ and hence $h = \frac{v_i^2}{2g}$ Now, in the kinematic equation of motion, $h = v_i t - \frac{g}{2}t^2$ hence $\frac{v_i^2}{2g} = v_it - \frac{g}{2}t^2$ which is equivalent to $\frac{g}{2}t^2 - v_it + \frac{v_i^2}{2g} = 0$ or in other words $t^2 - \frac{2v_i}{g} t + \frac{v_i^2}{g^2} = 0$ For this quadratic the discriminant is $\Delta = \frac{4v_i^2}{g^2} - \frac{4v_i^2}{g^2} = 0$ and therefore the (double) root is $t = \frac{v_i}{g}$ Hence the ability to make these equations consistent is completely independent of both the value of $$g$$ and the initial velocity $$v_i$$. That is just as well, as there is absolutely no physical reason why they shouldn't be. What I think you're doing wrong, @cinar, is using different values of g in both places. Pick any value you like, such as 9.8 m/s^2, find the corresponding value of h, and then find the time using the _same_ value of g.

25. cinar

I got it now because when I calculate h, I rounded it 11.5 but it s not h=11.479591836735

26. cinar

for h=11.479591836735 http://www.wolframalpha.com/input/?i=4.9t^2+-+15t+%2B+11.479591836735+%3D+0

27. cinar

I see the difference now for h=11.479591836734693877551020408163 http://www.wolframalpha.com/input/?i=4.9t^2+-+15t+%2B+11.479591836734693877551020408163%3D+0

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