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:

$y=h+ut+\frac{1}{2}at^2$

where

h = 23 m is the initial height

u = 27 m/s is the initial velocity, upward

$a=g=-9.8 m/s^2$ 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

$11=23+27t-4.9t^2\\4.9t^2-27t-12=0$

This is a second-order equation, which has solutions:

$t=\frac{27\pm \sqrt{(-27)^2-4(4.9)(-12)}}{2(4.9)}=\frac{27\pm \sqrt{964.2}}{9.8}$

$t_1 = -0.41 s$

$t_2=5.92 s$

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.

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8 0

3 years ago

What is the solution to the equation 5 (x minus 6) = 2 (x + 3)?

Reika [66]

The answer is x=12 :))

8 0

3 years ago

Need help with this pls I'm failing​

Need help with this pls I'm failing