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Physics 15 Online
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.

OpenStudy (mayankdevnani):

@RaphaelFilgueiras

OpenStudy (mayankdevnani):

@sauravshakya @ganeshie8 @Callisto @cinar

OpenStudy (mayankdevnani):

@.Sam. @hba

OpenStudy (mayankdevnani):

@Rohangrr @myininaya @amistre64 @Algebraic! @akash809

OpenStudy (mayankdevnani):

somebody.... solve it

OpenStudy (anonymous):

Use: v^2=u^2 +2as where u=0 , a=9.81m/s^2 , s=80m

OpenStudy (anonymous):

Now, can u?

OpenStudy (mayankdevnani):

i can , but can you solve it??

OpenStudy (mayankdevnani):

but , what is the value of time 't'

OpenStudy (mayankdevnani):

@RaphaelFilgueiras @Rohangrr

OpenStudy (anonymous):

U do not need t here..

OpenStudy (anonymous):

As height is known

OpenStudy (mayankdevnani):

why??

OpenStudy (mayankdevnani):

ok!! then

OpenStudy (anonymous):

v^2=u^2 +2as v^2=0^2+2*9.81*80 v^2=1569.6 v=39.6 m/s

OpenStudy (anonymous):

got it?

OpenStudy (mayankdevnani):

\[but, speed =\frac{distance}{time} \]

OpenStudy (anonymous):

That formula only works when the speed is constant

OpenStudy (mayankdevnani):

ok!

OpenStudy (anonymous):

When an object falls there is accleration

OpenStudy (anonymous):

|dw:1348062066625:dw| I need mass

OpenStudy (mayankdevnani):

total distance = 160m

OpenStudy (anonymous):

And for constant accleration there are some equation of motion.

OpenStudy (anonymous):

|dw:1348062114490:dw|

OpenStudy (mayankdevnani):

ya

OpenStudy (mayankdevnani):

v=u+at

OpenStudy (mayankdevnani):

\[or , a=(\frac{v-u}{t})\]

OpenStudy (anonymous):

Its undergoing free fall accelaration

OpenStudy (mayankdevnani):

ya

OpenStudy (anonymous):

Ya @myininaya

OpenStudy (anonymous):

opps

OpenStudy (anonymous):

Yep. @mayankdevnani

OpenStudy (mayankdevnani):

ok!!

OpenStudy (anonymous):

And also there is v^2=u^2+2as

OpenStudy (mayankdevnani):

we don't know acc. =?

OpenStudy (mayankdevnani):

we need to calculte the speed of body upon arrival on the ground

OpenStudy (anonymous):

wow.

OpenStudy (anonymous):

Accleration =acclration due to gravity=9.81m/s^2

OpenStudy (mayankdevnani):

but v =?

OpenStudy (mayankdevnani):

v= 0

OpenStudy (mayankdevnani):

right/wrong

OpenStudy (anonymous):

u=0

OpenStudy (anonymous):

screen-capped

OpenStudy (anonymous):

u need to calculate v

OpenStudy (mayankdevnani):

u= 0 why??

OpenStudy (mayankdevnani):

any reason?

OpenStudy (anonymous):

If U throw a ball upward with velocity v then ,|dw:1348062730065:dw|

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