Answer:
![3x_0](https://tex.z-dn.net/?f=3x_0)
Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
![d=v_x t](https://tex.z-dn.net/?f=d%3Dv_x%20t)
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
![d=v_0 t = x_0](https://tex.z-dn.net/?f=d%3Dv_0%20t%20%3D%20x_0)
in the second case, the horizontal velocity is increased to
![v_x' = 3v_0](https://tex.z-dn.net/?f=v_x%27%20%3D%203v_0)
And so the new distance travelled will be
![d' = v_x' t = 3 v_0 t = 3 x_0](https://tex.z-dn.net/?f=d%27%20%3D%20v_x%27%20t%20%3D%203%20v_0%20t%20%3D%203%20x_0)
So, the distance increases linearly with the horizontal velocity.
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
Answer:
The distance is 0.53 m.
Explanation:
Given that,
Target distance = 100.0 m
Speed of bullet = 300 m/s
We need to calculate the total time
Using formula of time
![t=\dfrac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bv%7D)
Put the value into the formula
![t=\dfrac{100.0}{300}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B100.0%7D%7B300%7D)
![t=0.33\ sec](https://tex.z-dn.net/?f=t%3D0.33%5C%20sec)
Now, consider vertical motion of bullet.
Initial velocity of bullet in vertical direction = 0 m/s
We need to calculate the vertically distance
Using equation of motion
![s=ut+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value in the equation
![s=0+\dfrac{1}{2}\times9.8\times(0.33)^2](https://tex.z-dn.net/?f=s%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%280.33%29%5E2)
![s=0.53\ m](https://tex.z-dn.net/?f=s%3D0.53%5C%20m)
Hence, The distance is 0.53 m.
<span>The answer is a heterogeneous mixture. Mixtures can be homogeneous and heterogeneous. If a solid and a liquid of a mixture cannot be separated and the difference between them is not visible, it is called homogeneous mixture (or solution). If a solid and a liquid of a mixture are visible and can be separated easily, the mixture is called heterogeneous.</span>