IIT JEE , AIEEE Forum - Physics , Mechanics

ruhi

hiiiiiiiiii...........................need help in this question..................

a drowsy cat spots a flowerpot that sails first up nd then down past an open window. the pot is in view 4 a total of 0.50 sec, and the top to bottom height of the window is 2m. how high above the window top does the flowerpot go??????

................................can someone explain me this question?????????

rajiv_nag

the answer should be 3.2mt.(g=10).

letA &B be the bottom & top of the window.
vel. at A=u
vel. at B=v.
time of travel from A to B will be (0.5)/2=0.25.
for motion from A to B ini & final vel. are knwn u&v.accel .is knwn.
t is knwn. we get,u+v=16.
& u-v=2.5
hence.v=8
for motion from B to max ht.,
ini & final vel are 8 & 0.accel. is -10.
we need s.
v^2 -u^2 =2as.
hence,s=3.2mt.
.

ruhi

hi
ans is 2.34 m
plzzzzzzz explain the question 2 me

AAKRITI

Ruhi, the question you have given is very elementary question of 1D free fall.

Let u be be the initial vel of the pot when it just enters the window view

now we are given the summed up time when the pot is in the scene, so for analysing motion of the pot when it enters the window view to when it leaves for achieving the max height, the time will be half of the total = 1/4 sec

s = ut + 1/2at² , take upward dirn as +y , s = +2m , a = -g, t = 1/4 sec

For the answer you have given, take g = 9.8m/s2

It gives u = 9.224m/s

Use this to analse the max height, H = u²/2g which comes out to be 4.34m

H = s + h { h is the height above the windoe that the pot goes }

This means h = 2.34 m

vikalp

i think akriti has done a gud job here .....

accept a thumz up from my side .........

........keeep it cooollll........

krishna.gopal

Good work Aakriti. Here i want to point a mistake by rajeev. Rajeev in u+v=16 and u-v=2.5 then how do you get v=8.

Please be careful about your calculations or you will get wrong answers with right approach

AAKRITI

thank you sir

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