IIT JEE , AIEEE Forum - Physics , Mechanics

sehgal

A stone is thrown vertically upwards from cliff with velocity 5m/s.it strikes the pond near the base of cliff after 4s.the height of cliff is:

a)6m b)60m c)40m d)100m.

give me the solution of this qn. Nd also tell me about the sign of u.also in other qn. I m not understanding the concept of sign of u nd v.,so tell me about this concept also

answer of this qn. Is b.

ronakaim

first using time of flight formula:
T-2u/g=1 secs.
this implies that the stone covers some height and comes back to where it was thrown from in 1 second. now from that point it is travelling under gravity with initial velocity 5 m/s( the stone aquires the same velocity of 5 as it was thrown with velocity 5)
H=UT+0.5*A*T*T
H=5*3+0.5*10*90
=60 MTS.

parismulye

yes ronakaim is right

nishant_88

height reached by stone is
h=u^2/2g=1.25m
time taken is
t=u/a=0.5s

Time taken to come to base t1=4-0.5=3.5s
total distance covered s=1/2g*(t1)^2=61.25m

So, height of cliff=61.25-1.25=60m

srkanubhav

time taken to reach the max. height = |u/a| = |5/-10| = 0.5 s
max. height reached = |u^2/2a| = 1.25m

now,
s=ut+1/2gt^2
= 1/2*10*12.25
= 61.25m

so the height of the cliff = 61.25-1.25= 60m

in my solution you will not have to worry about the sign of u and v
try to solve the other question by this method only.

DO RATE ME!!!!

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