A Ball is thrown vertically upwards.it was observed at a height h twice with a time interval  @t. the initial velocity of the ball is:

a) & 8gh+g^ (@t)^     b)& 8gh+(g@t/2)^        c)1/2  & 8gh+g^(@t)^     

d) & 8gh+4g^(@t)^.

Where   ^ =square ,     @=delta ,     &= underroot to all.  Please note in third option underroot is not present on ½   nd also note  in second option  2 is present upon g@t not only upon t.

PLEASE GIVE ME FULL SOLUTION OF THIS QN. ANSWER IS C.    

ans) let the ball be at a height h at time t n time t + t     

here h = ut - 1/2 gt² -----(1)  n h = u(t+t) - 1/2g(t+t)²     

equating  both of them v get t = 2u - gt/2g ------ (2)

subs (2) in (1) v get h = 4u² - g²(t)² / 8g 

therefore u = 1/2 

corrct me if wrong

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h=ut-1/2 gt^ (minus sign because velocity is upwards and acceleration is downwards)



gt^-2ut+2h=0



These gives two values of t and the difference between them is @t

(@t)^=(t1+t2)^-4*t1* t2

(@t)^=4*u^/g^-4*2h/g



This gives u=1/2 & 8hg+g^(@t)^



So answer is c

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