Question

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?

Answer 1

\[v^2=v_0^2-2gh\]

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

Answer 2

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

Answer 3

it is 2 times reach to apex

Answer 4

awesome! thanks

Answer 5

there is something wrong

Answer 6

what did you find t

Answer 7

1.5 s and total time 3 s

Answer 8

how about h?

Answer 9

is it 11.5

Answer 10

because t is not real, it is complex..

Answer 11

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

Answer 12

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..

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

maybe their teacher said them to take g=10

Answer 23

what is the formula use in pascal's law?

Answer 24

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.

Answer 25

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

h=11.479591836735

Answer 26

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

Answer 27

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