plss solve this question
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For the first phenomenon,
h=0.5gT²
For second phenomenon,
h`=0.5g(T/3)²
=>h`=(1/18)gT²
=>h`=(1/9)(0.5gT²)
=>h`=(1/9)h [From top]
now it will cover h-h` distance from ground.
h-h`=h-(1/9)h
=>h-h`=8h/9
Answer is C
h=0.5gT²
For second phenomenon,
h`=0.5g(T/3)²
=>h`=(1/18)gT²
=>h`=(1/9)(0.5gT²)
=>h`=(1/9)h [From top]
now it will cover h-h` distance from ground.
h-h`=h-(1/9)h
=>h-h`=8h/9
Answer is C
ravi123456789018th:
ok
in t seconds ball will cover distance = h metre. and ball will cover distance in t by 3 seconds = h/t multiply by t/3 after solving this we get distance = h/3 so distance from ground = h-h/3 = 2h/3
answer is 2h /3
is it correct
if not tell me the reason
no, you need not to apply arithmetical rules here since the two scenarios are not dependent on each other you have to apply falling body's equation two times for each of the incidence and then you can do the arithmetical calculations.
what is the problem in my method
it is unitary method
you're defying tge fact that height is proportional to the time squared by that way.
the*
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