Physics, 07.10.2019 21:20 Karlakins

Aball is tossed straight up into the air with an initial speed of 12 m/s.
a. how long does it take to reach its highest point?
b. how high does the ball rise above its point of release?
c. how long will it take for the ball to reach a point 5.0 m above its release
point? (should have two answers for this explain why)

Answers: 2

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Answer from: jaleesalewis2008

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:

v_f = a(t_t) + v_i

and do some algebra to get:

$\frac{v_f - v_i}{a} = t_t$

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):

$\frac{0- 12m/s}{-gm/s^2} = t_t$

$\frac{-12m/s}{-9.8m/s^2} = t_t$

$\frac{12}{9.8s} = t_t$

$1.22449s \approx t_t$

To find the final height we can use:

$x_f = \frac{v^2_f - v^2_i}{2a}$

and plug in our values to get:

$x_f = \frac{0^2m/s - 12^2m^2s^2}{-2g}$

$x_f = \frac{-144m^2s^2}{-2 \cdot 9.8m/s^2}$

$x_f = \frac{-144m^2s^2}{-19.6m/s^2}$

$x_f = \frac{144m}{19.6}$

$x_f \approx 14.69387m$

To find the time we can use the time dependent position equation:

$x_f = a(\frac{t^2_t}{2}) + v_i(t_t) + x_i$

This here can be made into a quadratic equation like so (xi is set up to be 0m, so the equation wont have it):

at^2_t} + 2v_it_t - 2x_f = 0

Here we can use the quadratic formula:

$t_t = \frac{-2v_i \pm \sqrt{(2v_i)^2 - 4(a)(-2x_f)} }{2(a)}$

And now it would be best if we put in our values (xf = 5m because that is our question):

$t_t = \frac{-2(12m/s) \pm \sqrt{(2(12m/s))^2 - 4(-9.8m/s^2)(-2(5m))} }{2(-9.8m/s^2)}$

$t_t = \frac{(-24m/s) \pm \sqrt{(24^2m^2/s^2) + (39.2m/s^2)(-10m)} }{(-19.6m/s^2)}$

$t_t = \frac{(-24m/s) \pm \sqrt{(576m^2/s^2) + (-392m^2/s^2)} }{(-19.6m/s^2)}$

$t_t = \frac{(-24m/s) \pm \sqrt{(184m^2/s^2)} }{(-19.6m/s^2)}$

$t_t = \frac{(-24m/s) \pm (\sqrt{184}) m/s}{(-19.6m/s^2)}$

Finally we have simplified enough to be worth solving for:

$t_t = \frac{(-24) \pm (\sqrt{184})}{(-19.6s)}$

We get:

$t_t \approx 0.532415s \approx \frac{(-24) + (\sqrt{184})}{(-19.6s)}$

and

$t_t \approx 1.916564s \approx \frac{(-24) - (\sqrt{184})}{(-19.6s)}$

Because time is always positive we want to choose the plus answer.

Answer from: Quest

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Aball is tossed straight up into the air with an initial speed of 12 m/s.
a. how long does it t...

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