anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
I found the total time it takes the ball to reach the apex but can't find the total time in flight?
anonymous
  • anonymous
it is 2 times reach to apex

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
awesome! thanks
anonymous
  • anonymous
there is something wrong
anonymous
  • anonymous
what did you find t
anonymous
  • anonymous
1.5 s and total time 3 s
anonymous
  • anonymous
how about h?
anonymous
  • anonymous
is it 11.5
anonymous
  • anonymous
because t is not real, it is complex..
anonymous
  • anonymous
its initial speed is 15 right, can you check it again?
anonymous
  • 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
  • anonymous
but I'am still curies why that eq. did not work
JamesJ
  • 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
  • JamesJ
hence it's at the apex when v(t) = 0.
anonymous
  • anonymous
what if I want to solve problem using by those formulas, it must be worked, where am I making wrong?
JamesJ
  • 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
  • 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
  • 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
  • anonymous
that's interesting when I accept g=9.8 it is not real
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=4.9t^2+-+15t+%2B+11.5+%3D+0
anonymous
  • anonymous
maybe their teacher said them to take g=10
anonymous
  • anonymous
what is the formula use in pascal's law?
JamesJ
  • 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
  • anonymous
I got it now because when I calculate h, I rounded it 11.5 but it s not h=11.479591836735
anonymous
  • anonymous
for h=11.479591836735 http://www.wolframalpha.com/input/?i=4.9t^2+-+15t+%2B+11.479591836735+%3D+0
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.