Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
Explanation and Examples
let the mass of the compressor be
mass (m):
height in x axis is (h1)
height in y axis be (h2):
Height difference: h2-h1
displacement x force:
mass x gravity x height
(m)*9.8*(height difference) = ___ J
Since gravity is forcing down, it would be negative!
Put the values that you require and get the answer.
<h3><u>Answer;</u></h3>
Large mirrors are easier to build than large lenses.
<h3><u>Explanation;</u></h3>
- <em><u>Reflector telescopes have a number of advantages as compared to refracting telescopes and other types of telescopes. </u></em>
- <em><u>Reflector telescopes do not suffer from chromatic aberration because all wavelengths will reflect off the mirror in the same way. The support for the objective mirror is all along the back side so they can be made very large.</u></em>
- Additionally, reflector telescopes are cheaper to make than refractors of the same size. Also since in reflector telescopes light is reflecting off the objective, rather than passing through it, only one side of the reflector telescope's objective needs to be perfect.
Answer:
Explanation:
Horizontal displacement
x = 120 t
Vertical position
y = 3610 - 4.9 t²
y = 0 for the ground
0 = 3610 - 4.9 t²
t = 27.14 s
This is the time it will take to reach the ground .
During this period , horizontal displacement
x = 120 x 27.14 m
= 3256.8 m
So packet should be released 3256.8 m before the target.