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catsrule332
Group Title
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?
 2 years ago
 2 years ago
catsrule332 Group Title
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?
 2 years ago
 2 years ago

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cinar Group TitleBest ResponseYou've already chosen the best response.2
\[v^2=v_0^22gh\] v=0 v_0=15 find h then use this formula to find out t \[h=v _{i}t\frac12g t^2\]
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
I found the total time it takes the ball to reach the apex but can't find the total time in flight?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
it is 2 times reach to apex
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
awesome! thanks
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
there is something wrong
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
what did you find t
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
1.5 s and total time 3 s
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
because t is not real, it is complex..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
its initial speed is 15 right, can you check it again?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
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..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
but I'am still curies why that eq. did not work
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
hence it's at the apex when v(t) = 0.
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
what if I want to solve problem using by those formulas, it must be worked, where am I making wrong?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
no it is complex, and when I put t=1.5 right side is 0 left side is 11.25 and also \[\Delta <0\]
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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 \[ \Delta = (15)^2  4(5)(11.25) = 0 \]
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
that's interesting when I accept g=9.8 it is not real
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
http://www.wolframalpha.com/input/?i=4.9t^2++15t+%2B+11.5+%3D+0
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
maybe their teacher said them to take g=10
 2 years ago

chuchay Group TitleBest ResponseYou've already chosen the best response.0
what is the formula use in pascal's law?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
I got it now because when I calculate h, I rounded it 11.5 but it s not h=11.479591836735
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
for h=11.479591836735 http://www.wolframalpha.com/input/?i=4.9t^2++15t+%2B+11.479591836735+%3D+0
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.2
I see the difference now for h=11.479591836734693877551020408163 http://www.wolframalpha.com/input/?i=4.9t^2++15t+%2B+11.479591836734693877551020408163%3D+0
 2 years ago
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