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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- anonymous

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

- anonymous

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

- anonymous

it is 2 times reach to apex

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- anonymous

awesome! thanks

- anonymous

there is something wrong

- anonymous

what did you find t

- anonymous

1.5 s and total time 3 s

- anonymous

how about h?

- anonymous

is it 11.5

- anonymous

because t is not real, it is complex..

- anonymous

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

- anonymous

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..

- anonymous

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

- 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

- JamesJ

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

- anonymous

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

- 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

- anonymous

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

- 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
\[ \Delta = (-15)^2 - 4(5)(11.25) = 0 \]

- anonymous

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

- anonymous

http://www.wolframalpha.com/input/?i=4.9t^2+-+15t+%2B+11.5+%3D+0

- anonymous

maybe their teacher said them to take g=10

- anonymous

what is the formula use in pascal's law?

- 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.

- anonymous

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

- anonymous

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

- anonymous

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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