Physics: Astone is thrown vertically upward with a speed of 24.0 m/s. (a) how fast is it moving when it is at a height of 13.0 m? (b) how much time is required to reach this height? (c) why are there two answers to (b)?

Astone is thrown vertically upward with a speed - id2787832, Terezagonzalez2271

hahHcjckk

04.01.2022

Explanation:

initial speed, u = 24 m/s

acceleration due to gravity, g = - 9.8 m/s^2

(a) h = 13 m

Let it is moving with velocity v at a height of 13 m.

Use third equation of motion

$v^{2}=u^{2}+2gh$

By substituting the values

$v^{2}=24^{2}- 2 \times 9.8 \times 13$

v = 17.9 m/s

(b) Let it takes time t to reach at height 13 m

Use second equation of motion

$s=ut+\frac{1}{2}gt^{2}$

$13=24t-4.9t^{2}$

$4.9t^{2}-24t+13=0$

$t=\frac{24\pm \sqrt{24^{2}-4\times4.9\times13}}{2\times 4.9}$

$t=\frac{24 \pm 17.92}{9.8$

t = 4.28 second or t = 0.62 second

(c) the value of t = 0.62 second shows that the journey is upward and at t = 4.28 second, the stone is in downward journey.

0,0(0 оценок)