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.
13 years ago
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OpenStudy (mayankdevnani):
@RaphaelFilgueiras
13 years ago
OpenStudy (mayankdevnani):
@sauravshakya @ganeshie8 @Callisto @cinar
13 years ago
OpenStudy (mayankdevnani):
@.Sam. @hba
13 years ago
OpenStudy (mayankdevnani):
@Rohangrr @myininaya @amistre64 @Algebraic! @akash809
13 years ago
OpenStudy (mayankdevnani):
somebody.... solve it
13 years ago
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OpenStudy (anonymous):
Use:
v^2=u^2 +2as
where u=0 , a=9.81m/s^2 , s=80m
13 years ago
OpenStudy (anonymous):
Now, can u?
13 years ago
OpenStudy (mayankdevnani):
i can , but can you solve it??
13 years ago
OpenStudy (mayankdevnani):
but , what is the value of time 't'
13 years ago
OpenStudy (mayankdevnani):
@RaphaelFilgueiras @Rohangrr
13 years ago
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OpenStudy (anonymous):
U do not need t here..
13 years ago
OpenStudy (anonymous):
As height is known
13 years ago
OpenStudy (mayankdevnani):
why??
13 years ago
OpenStudy (mayankdevnani):
ok!! then
13 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
13 years ago
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OpenStudy (anonymous):
got it?
13 years ago
OpenStudy (mayankdevnani):
\[but, speed =\frac{distance}{time} \]
13 years ago
OpenStudy (anonymous):
That formula only works when the speed is constant
13 years ago
OpenStudy (mayankdevnani):
ok!
13 years ago
OpenStudy (anonymous):
When an object falls there is accleration
13 years ago
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OpenStudy (anonymous):
|dw:1348062066625:dw| I need mass
13 years ago
OpenStudy (mayankdevnani):
total distance = 160m
13 years ago
OpenStudy (anonymous):
And for constant accleration there are some equation of motion.
13 years ago
OpenStudy (anonymous):
|dw:1348062114490:dw|
13 years ago
OpenStudy (mayankdevnani):
ya
13 years ago
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OpenStudy (mayankdevnani):
v=u+at
13 years ago
OpenStudy (mayankdevnani):
\[or , a=(\frac{v-u}{t})\]
13 years ago
OpenStudy (anonymous):
Its undergoing free fall accelaration
13 years ago
OpenStudy (mayankdevnani):
ya
13 years ago
OpenStudy (anonymous):
Ya @myininaya
13 years ago
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OpenStudy (anonymous):
opps
13 years ago
OpenStudy (anonymous):
Yep. @mayankdevnani
13 years ago
OpenStudy (mayankdevnani):
ok!!
13 years ago
OpenStudy (anonymous):
And also there is v^2=u^2+2as
13 years ago
OpenStudy (mayankdevnani):
we don't know acc. =?
13 years ago
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OpenStudy (mayankdevnani):
we need to calculte the speed of body upon arrival on the ground
13 years ago
OpenStudy (anonymous):
wow.
13 years ago
OpenStudy (anonymous):
Accleration =acclration due to gravity=9.81m/s^2
13 years ago
OpenStudy (mayankdevnani):
but v =?
13 years ago
OpenStudy (mayankdevnani):
v= 0
13 years ago
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OpenStudy (mayankdevnani):
right/wrong
13 years ago
OpenStudy (anonymous):
u=0
13 years ago
OpenStudy (anonymous):
screen-capped
13 years ago
OpenStudy (anonymous):
u need to calculate v
13 years ago
OpenStudy (mayankdevnani):
u= 0 why??
13 years ago
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OpenStudy (mayankdevnani):
any reason?
13 years ago
OpenStudy (anonymous):
If U throw a ball upward with velocity v then ,|dw:1348062730065:dw|
13 years ago