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CBSE Class 11-science Answered

please solve

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Asked by prakriti12oct | 04 Nov, 2020, 01:52: AM

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Let u be the projection velocity of ball at the point of hitting the ball . Let θ be the projection angle.

After 4 second , velocity of ball v is given as,

v = ( u sinθ ) - ( g t ) = ( u sinθ ) - ( 4 × 9.8 ) m/s = [ ( u sinθ ) - 39.2 ] m/s

Horizontal distance = ( u cosθ) × 4 = 60 , hence horizontal component of velocity = u cosθ = ( 60 / 4 ) = 15 m/s ...............(1)

If the direction of velocity of ball makes angle 45^o to the horizontal while moving downward, then we have

begin mathsize 14px style fraction numerator V e r t i c a l space c o m p o n e n t space o f space v e l o c i t y over denominator H o r i z o n t a l space c o m p o n e n t space o f space v e l o c i t y end fraction space equals space minus tan 45 space equals space minus 1 end style

Hence we get, [ ( u sinθ ) - 39.2 ] = - 15 or u sinθ = 39.2 -15 = 24.2 .................(2)

By dividing eqn.(2) by eqn.(1) , we get , tanθ = 24.2/15 = 1.613 or θ = 58^o

By substituting for θ, we get from eqn.(1) , u = 24.2 / sin(58) ≈ 28.5 m/s

since from the list of options ,

begin mathsize 14px style square root of 850 end style

is very near to u = 28.5 m/s , we choose that option

begin mathsize 14px style square root of 850 end style

m/s as the answer

Answered by Thiyagarajan K | 04 Nov, 2020, 09:22: AM

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