Answer:
a. ≈ 1.22449s
b. ≈ 14.69387m
c. ≈ 0.532415s
Explanation:
Because we are trying to find when it is at it's highest point we can safely say that it's velocity at that point is 0m/s,
therefore we can use the equation:
[tex]v_f = a(t_t) + v_i[/tex]
and do some algebra to get:
[tex]\frac{v_f - v_i}{a} = t_t[/tex]
Now we plug in our values (note that this is assumed to be on Earth and that because it says that upwards is positive, we set g to be negative to say that it is pointing down):
[tex]\frac{0- 12m/s}{-gm/s^2} = t_t[/tex]
[tex]\frac{-12m/s}{-9.8m/s^2} = t_t[/tex]
[tex]\frac{12}{9.8s} = t_t[/tex]
[tex]1.22449s \approx t_t[/tex]
To find the final height we can use:
[tex]x_f = \frac{v^2_f - v^2_i}{2a}[/tex]
and plug in our values to get:
[tex]x_f = \frac{0^2m/s - 12^2m^2s^2}{-2g}[/tex]
[tex]x_f = \frac{-144m^2s^2}{-2 \cdot 9.8m/s^2}[/tex]
[tex]x_f = \frac{-144m^2s^2}{-19.6m/s^2}[/tex]
[tex]x_f = \frac{144m}{19.6}[/tex]
[tex]x_f \approx 14.69387m[/tex]
To find the time we can use the time dependent position equation:
[tex]x_f = a(\frac{t^2_t}{2}) + v_i(t_t) + x_i[/tex]
This here can be made into a quadratic equation like so (xi is set up to be 0m, so the equation wont have it):
[tex]at^2_t} + 2v_it_t - 2x_f = 0[/tex]
Here we can use the quadratic formula:
[tex]t_t = \frac{-2v_i \pm \sqrt{(2v_i)^2 - 4(a)(-2x_f)} }{2(a)}[/tex]
And now it would be best if we put in our values (xf = 5m because that is our question):
[tex]t_t = \frac{-2(12m/s) \pm \sqrt{(2(12m/s))^2 - 4(-9.8m/s^2)(-2(5m))} }{2(-9.8m/s^2)}[/tex]
[tex]t_t = \frac{(-24m/s) \pm \sqrt{(24^2m^2/s^2) + (39.2m/s^2)(-10m)} }{(-19.6m/s^2)}[/tex]
[tex]t_t = \frac{(-24m/s) \pm \sqrt{(576m^2/s^2) + (-392m^2/s^2)} }{(-19.6m/s^2)}[/tex]
[tex]t_t = \frac{(-24m/s) \pm \sqrt{(184m^2/s^2)} }{(-19.6m/s^2)}[/tex]
[tex]t_t = \frac{(-24m/s) \pm (\sqrt{184}) m/s}{(-19.6m/s^2)}[/tex]
Finally we have simplified enough to be worth solving for:
[tex]t_t = \frac{(-24) \pm (\sqrt{184})}{(-19.6s)}[/tex]
We get:
[tex]t_t \approx 0.532415s \approx \frac{(-24) + (\sqrt{184})}{(-19.6s)}[/tex]
and
[tex]t_t \approx 1.916564s \approx \frac{(-24) - (\sqrt{184})}{(-19.6s)}[/tex]
Because time is always positive we want to choose the plus answer.
