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