Physics
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OpenStudy (mayankdevnani):
A ball thrown vertically upward reaches a height of 80 m . Calculate the speed of the ball upon arrival on the ground.
Also Calculate the time to reach the highest point.
14 years ago
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OpenStudy (mayankdevnani):
@RaphaelFilgueiras
14 years ago
OpenStudy (mayankdevnani):
@sauravshakya @ganeshie8 @Callisto @cinar
14 years ago
OpenStudy (mayankdevnani):
@.Sam. @hba
14 years ago
OpenStudy (mayankdevnani):
@Rohangrr @myininaya @amistre64 @Algebraic! @akash809
14 years ago
OpenStudy (mayankdevnani):
somebody.... solve it
14 years ago
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OpenStudy (anonymous):
Use:
v^2=u^2 +2as
where u=0 , a=9.81m/s^2 , s=80m
14 years ago
OpenStudy (anonymous):
Now, can u?
14 years ago
OpenStudy (mayankdevnani):
i can , but can you solve it??
14 years ago
OpenStudy (mayankdevnani):
but , what is the value of time 't'
14 years ago
OpenStudy (mayankdevnani):
@RaphaelFilgueiras @Rohangrr
14 years ago
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OpenStudy (anonymous):
U do not need t here..
14 years ago
OpenStudy (anonymous):
As height is known
14 years ago
OpenStudy (mayankdevnani):
why??
14 years ago
OpenStudy (mayankdevnani):
ok!! then
14 years ago
OpenStudy (anonymous):
v^2=u^2 +2as
v^2=0^2+2*9.81*80
v^2=1569.6
v=39.6 m/s
14 years ago
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OpenStudy (anonymous):
got it?
14 years ago
OpenStudy (mayankdevnani):
\[but, speed =\frac{distance}{time} \]
14 years ago
OpenStudy (anonymous):
That formula only works when the speed is constant
14 years ago
OpenStudy (mayankdevnani):
ok!
14 years ago
OpenStudy (anonymous):
When an object falls there is accleration
14 years ago
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OpenStudy (anonymous):
|dw:1348062066625:dw| I need mass
14 years ago
OpenStudy (mayankdevnani):
total distance = 160m
14 years ago
OpenStudy (anonymous):
And for constant accleration there are some equation of motion.
14 years ago
OpenStudy (anonymous):
|dw:1348062114490:dw|
14 years ago
OpenStudy (mayankdevnani):
ya
14 years ago
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OpenStudy (mayankdevnani):
v=u+at
14 years ago
OpenStudy (mayankdevnani):
\[or , a=(\frac{v-u}{t})\]
14 years ago
OpenStudy (anonymous):
Its undergoing free fall accelaration
14 years ago
OpenStudy (mayankdevnani):
ya
14 years ago
OpenStudy (anonymous):
Ya @myininaya
14 years ago
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OpenStudy (anonymous):
opps
14 years ago
OpenStudy (anonymous):
Yep. @mayankdevnani
14 years ago
OpenStudy (mayankdevnani):
ok!!
14 years ago
OpenStudy (anonymous):
And also there is v^2=u^2+2as
14 years ago
OpenStudy (mayankdevnani):
we don't know acc. =?
14 years ago
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OpenStudy (mayankdevnani):
we need to calculte the speed of body upon arrival on the ground
14 years ago
OpenStudy (anonymous):
wow.
14 years ago
OpenStudy (anonymous):
Accleration =acclration due to gravity=9.81m/s^2
14 years ago
OpenStudy (mayankdevnani):
but v =?
14 years ago
OpenStudy (mayankdevnani):
v= 0
14 years ago
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OpenStudy (mayankdevnani):
right/wrong
14 years ago
OpenStudy (anonymous):
u=0
14 years ago
OpenStudy (anonymous):
screen-capped
14 years ago
OpenStudy (anonymous):
u need to calculate v
14 years ago
OpenStudy (mayankdevnani):
u= 0 why??
14 years ago
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OpenStudy (mayankdevnani):
any reason?
14 years ago
OpenStudy (anonymous):
If U throw a ball upward with velocity v then ,|dw:1348062730065:dw|
14 years ago