IIT JEE , AIEEE Forum - Physics , Mechanics

Anupamagrawal13a

Ques-A person sitting on the top of a building is dropping balls at regular intervals of one second.find the positions of the third,4th,5th ball when the 6th ball is dropped.

Ans-44.1m,19.6m,4.9m below the top

Please answer it quickly.

netkid07

taking downward direction as +ve

we know that

s=ut+1/2at^2

u=0
a=g

s=1/2gt^2

or

s=4.9 t^2.........(x)

put the values of t for diff balls
difference b/w thrownig of 6th and 5th ball is 1 sec

put in (x) u get 4.9 m

difference b/w thrownig of 6th and 4th ball is 2 sec

put in (x) u get 19.6 m

similarly do other one

nudge if u hav doubt
rate if u like !!

iitkgp_bipin

When 6th ball is dropped 5th, 4th and 3rd balls have travelled for 1 s, 2 s and 3 s respectively.

h = ut + (1/2)gt²

since balls are dropped u=0 and we know g = 9.8 m/s²

h = 4.9t²

For 5th ball : t = 1
h = 4.9 m

For 4th ball : t = 2
h = 4.9(2²) = 19.6 m

For 3rd ball : t = 3
h = 4.9(3²) = 44.1 m

netkid07

bipin

i think g=9.8m/s^2 not 4.9 m/s^2

learner

dude he's done g/2....he's right!!

Anupamagrawal13a

Bipin Sir gave the correct answer.rates to you all for the help.

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