Science, asked by chiragverma, 1 year ago

A ball is released from the top of a tower of height h matres. it takes T seconds to reach the ground. what is the position of the ball in T/3 seconds?

Answers

Answered by abhi178

when ball released from the tower then initial velocity =0
at T/3 time velocity of ball (v)
v=u+at
v=0-gT/3
=-gT/3 m/s
now position of ball from top of tower is ,
v^2=u^2+2aS
(-gT/3)^2=0 -2g (-h")
h"=gT^2/18 ------------(1)

but we also know
h=1/2gT^2
hence T^2=2h/g
put T^2 value in equation (1)
h"=g (2h/g)/18=h/9
hence position of object from ground is h-h/9=8h/9

abhi178: please mark as brainliest

abhi178: am I right ?

chiragverma: no

abhi178: then what is answer

chiragverma: answer will be (8h/9)m above the ground

abhi178: dud I am correct this is position from ground of tower but my top of tower

chiragverma: please tell me accurate answer

abhi178: now you look answer

abhi178: if satisfied please mark as brainliest

chiragverma: thanks

Answered by Satwatneyearthian

this is answer to ur question

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Satwatneyearthian: Erandol , Jalgaon , Maharashtra

chiragverma: please check your answer

Satwatneyearthian: i know

Satwatneyearthian: it is wrong

chiragverma: what do you

Satwatneyearthian: ???? what do you????

chiragverma: means aap kya karte h

Satwatneyearthian: 10th standard

chiragverma: OK I read in ninth standard

chiragverma: do you know holy mother English school

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