the rock will be at 11 meters from the ground level after 5.92 seconds
Step-by-step explanation:
The motion of the rock is a free-fall motion, since the rock is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, so its position at time t is given by the equation:
![y=h+ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dh%2But%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
h = 23 m is the initial height
u = 27 m/s is the initial velocity, upward
is the acceleration of gravity, downward
t is the time
We want to find the time t at which the position of the rock is
y = 11 m
Substituting and re-arranging the equation, we find
![11=23+27t-4.9t^2\\4.9t^2-27t-12=0](https://tex.z-dn.net/?f=11%3D23%2B27t-4.9t%5E2%5C%5C4.9t%5E2-27t-12%3D0)
This is a second-order equation, which has solutions:
![t=\frac{27\pm \sqrt{(-27)^2-4(4.9)(-12)}}{2(4.9)}=\frac{27\pm \sqrt{964.2}}{9.8}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B27%5Cpm%20%5Csqrt%7B%28-27%29%5E2-4%284.9%29%28-12%29%7D%7D%7B2%284.9%29%7D%3D%5Cfrac%7B27%5Cpm%20%5Csqrt%7B964.2%7D%7D%7B9.8%7D)
So
![t_1 = -0.41 s](https://tex.z-dn.net/?f=t_1%20%3D%20-0.41%20s)
![t_2=5.92 s](https://tex.z-dn.net/?f=t_2%3D5.92%20s)
The first solution is negative so we neglect it: therefore, the rock will be at 11 meters from the ground level after 5.92 seconds.
Learn more about free fall:
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