NEET Class neet Answered

Initially, ball is thrown vertically upward with speed u . Maximum height reached by the ball is  [ u² / (2g) ]

Hence total distance travelled before first bounce is ( u²/ g ) .

After first bouncing , speed of ball is ( e u ) .

Hence total distance travelled before second bounce is ( e² u² ) /  g  .

After second bouncing , speed of ball is ( e² u ) .

Hence total distance travelled before third bounce is ( e⁴ u² ) /  g .

................so on

Hence total distance D travelled by the ball is

D = ( u² / g ) [ 1 / (1 - e ) ]

Let us substitute the values u = 40 m/s , g = 10 m/s² and e = 0.5 , then we get

D = ( 1600 / 10 ) [ 1 / (1-0.5) ] = 320 m

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Initially, ball is thrown vertically upward with speed u .  Time taken by the ball to reach maximum height is  [ u / g ]

Hence total time travelled before first bounce is [ ( 2u ) / g ] .

After first bouncing , speed of ball is ( e u ) .

Hence total time travelled before second bounce is ( 2 e u ) /  g .

After second bouncing , speed of ball is ( e² u ) .

Hence total time travelled before third bounce is ( 2 e² u ) /  g .

............so on

Hence total time T taken by ball during bouncing is

T = [ (2u)/g ] [ 1 + e + e² .................... ]

T = [ (2u)/g ] [ 1/(1-e) ]  = [ ( 2 × 40 ) /10 ] [ 1 / (1-0.5) ] = 16 s