1.

A ball thrown up is caught by the thrower after 6s after start. the height of the ball has risen is (g=10m/s2

Answer»

Hello friend..!!!

in the above question , we should calculate the height attained by the ball ,

we know H = ut - 1/2gt²

u = ?? 

t = 6 sec 

h = ??

we know that ,

v = u + at  ( but here v = -u before reaching the ground )

-u = u + at 

-2u = at  ( a = g = 9.8 and t = 6 )

u = ( 9.8 x 6 ) / 2 

u = 29 . 4 m / s -----------------------( 1 ) 

we know,

H = ut - 1/2 gt²

H = (29.4)(3) - 1/2(9.8)(3X 3 )

here i considered t = 3 sec because , the question asked is for max height so if the TOTAL time taken for the journey is 6 sec then half of it is 3 sec.

H = 44.11 m

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hope it helps...!!


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