The greatest height the ball will attain is 3.27 m
<h3>Data obtained from the question</h3>
- Initial velocity (u) = 8 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the ball can attain can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 8² – (2 × 9.8 × h)
0 = 64 – 19.6h
Collect like terms
0 – 64 = –19.6h
–64 = –19.6h
Divide both side by –19.6
h = –64 / –19.6h
h = 3.27 m
Thus, the greatest height the ball can attain is 3.27 m
Learn more about motion under gravity:
brainly.com/question/13914606
Answer:
The duration is ![T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m](https://tex.z-dn.net/?f=T%20%20%3D72%20%5C%20%20years%20%2Ftex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EFrom%20the%20question%20we%20are%20told%20that%20%3C%2Fp%3E%3Cp%3E%20%20%20%20The%20%20distance%20is%20%20%5Btex%5DD%20%20%3D%20%2035%20%5C%20light-years%20%3D%2035%20%2A%20%209.46%20%2A10%5E%7B15%7D%20%3D%203.311%20%2A10%5E%7B17%7D%20%5C%20%20m%20)
Generally the time it would take for the message to get the the other civilization is mathematically represented as

Here c is the speed of light with the value 
=> 
=> 
converting to years



Now the total time taken is mathematically represented as

=> 
=> [tex]T =72 \ years /tex]
Answer:
Explanation:
Given
initial velocity component of engines is


time period of engine running=763 s
Displacement in 

Using
in x and y direction




In y direction




x component
y component
Elements that give up electrons easily are called <u>metals.</u>
hope this helps!