The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;
![h = v_0_yt + \frac{1}{2} gt^2](https://tex.z-dn.net/?f=h%20%3D%20v_0_yt%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>
![h = \frac{1}{2} gt^2](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
The time taken by the first stone to hit the ground is calculated as;
![t_1 = \sqrt{\frac{2h}{g} }](https://tex.z-dn.net/?f=t_1%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D)
When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as
![t_2 = \sqrt{\frac{2h}{g} } + 1](https://tex.z-dn.net/?f=t_2%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D%20%2B%201)
![t_2 = t_1 + 1](https://tex.z-dn.net/?f=t_2%20%3D%20t_1%20%2B%201)
Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
Learn more here:brainly.com/question/16793944
Answer:
The initial velocity is 12.34[m/s]
Explanation:
To understand this problem, let's draw a sketch of what is happening. In the attached image we can fin de sketch.
According to the equations of uniform movement accelerated we have to express them in the two components x and y.
In this problem we will analyze first the movement in the y axis and its equation to obtain the velocity in the initial point (o) and then with the angle of 33° respect to the horizontal we can find the main velocity.
![v_{y}^{2} =(v_{y})_{o}^{2} -2*a*y\\where:\\v_{y}=velocity in the -y- component \\a=gravity acceleration [m/s^2]\\y=maximum heigth [m]\\](https://tex.z-dn.net/?f=v_%7By%7D%5E%7B2%7D%20%20%3D%28v_%7By%7D%29_%7Bo%7D%5E%7B2%7D%20%20-2%2Aa%2Ay%5C%5Cwhere%3A%5C%5Cv_%7By%7D%3Dvelocity%20in%20the%20-y-%20component%20%5C%5Ca%3Dgravity%20acceleration%20%5Bm%2Fs%5E2%5D%5C%5Cy%3Dmaximum%20heigth%20%5Bm%5D%5C%5C)
In the previous equation we have the negative term - (2*a*y), the reason for the negative sign is because gravity is the acceleration that is acting downwards.
Therefore:
![(0)^{2} =(v_{y})_{o}^{2} -2*9.81*2.3\\\\(v_{y})_{o} = \sqrt{2*9.81*2.3} \\(v_{y})_{o} =6.72[m/s]](https://tex.z-dn.net/?f=%280%29%5E%7B2%7D%20%20%20%3D%28v_%7By%7D%29_%7Bo%7D%5E%7B2%7D%20%20-2%2A9.81%2A2.3%5C%5C%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%20%5Csqrt%7B2%2A9.81%2A2.3%7D%20%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%3D6.72%5Bm%2Fs%5D)
Now with the velocity in the y component, we can calculate the main vector of the velocity.
![(v_{y})_{o}=v*sin(33)\\v=\frac{(v_{y})_{o}}{sin(33)}\\ v=12.33[m/s]](https://tex.z-dn.net/?f=%28v_%7By%7D%29_%7Bo%7D%3Dv%2Asin%2833%29%5C%5Cv%3D%5Cfrac%7B%28v_%7By%7D%29_%7Bo%7D%7D%7Bsin%2833%29%7D%5C%5C%20v%3D12.33%5Bm%2Fs%5D)
The main topic to understand in this problem is that when the height is maximum the velocity will be zero in the y component, therefore Vy=0 and we can simplify the problem.