A space shuttle requires solid rocket boosters to be lifted into orbit in lower space. At certain altitude, the boosters are separated from the liquid fuel tank and the fuel outputs are cut off. if the separation occurs at 45 km above sea level and upward is the positive direction, what is the VELOCITY of the boosters immediately before they strike the ocean surface?
@zepdrix please help me :D
@thomaster
@Abhisar
Hello @BestRivenNA !
hi
Have you tried it urself ?
ya
i thought that the problem didnt have enough information
It has given you !
ummmm given me??
Do you know the principal of conservation of mechanical energy ?
nope
Then chuck it !
...
uhh sure
\(\color{blue}{\text{Originally Posted by}}\) @Abhisar Do you know Newton's Second Equation of motion ? \(\color{blue}{\text{End of Quote}}\)
S=ut + 1/2at^2 ?
i dont know what s and u stand for
No u will have to use third equation of motion, \(\sf v^2=u^2+2as\) V= final velocity, u=initial velocity, s=distance, a=acceleration.
oh yeah i know that one
Since, the boosters are falling from 45km Take S=4500 m u=0 a=9.8 and find the value of v
but how do you find the acceleration without being given the velocity?
@Abhisar help pls :D
??
acceleration due to gravity is fixed. It's value is 9.8 m/s^2
ohhhhhhhhh yeah thanks!
u got the answers ?
and do you know why u=0 ?
well i got 297 m/s
uhh and yeah i dont really know why u=0
\(\color{blue}{\text{Originally Posted by}}\) @BestRivenNA well i got 297 m/s \(\huge\checkmark\) \(\color{blue}{\text{End of Quote}}\)
Whenever something `Freely Falls` under gravity or is `Dropped`, the initial velocity u=0
ok thanks for your AWESOME help!
\(\color{red}{\huge\bigstar}\huge\text{You're Most Welcome! }\color{red}\bigstar\) \(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)
lol
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