Question

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 1

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...!!

Answer 2

Answer:

45 Meters

Explanation:

From the given Question,

=>Ball returns to the same position after 6 seconds from throwing...

=>Time of ascend=Time of descent

So, time (t) = 3s for both time of ascend and also for decent

So, by the formula

v=u-gt,

=>0=u-gt,

=u-10x3

=30 m/s =>speed of the ball

Height the ball has risen=> by, ut- 1/2 gt^2

=(30x3)-0.5x10x3^2

=90-45. = 45 meters

#Hope_It_Helps