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A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t =0 to 12 s. |
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Answer» Solution :Here initial speed `v_(0)=0 h = 90 m` Lost speed in each collision, `=1/10 xx ` Speed acquired while striking ground, If the velocity of ball while striking ground is `v_(1),` then `V _(1) ^(2) - v _(0) ^(2) = 2gh` `v _(1) ^(2-) -0 = 2 xx 9.8 xx 90` `v _(1) ^(2) sqrt (2 xx 9.8 xx 90)` ` v _(1) = 42 ms ^(-1)` and time taken to reach the ground is `t_(1), `then From `d = v_(0) t + 1/2 G t ^(2) ` `h = 1/2 g t _(1) ^(2)` `t _(1) = sqrt ((2g)/(g))` ` = sqrt ((2 xx 90)/(9.8))` `t _(1) = 4.28 s` If the velocity when returning back after striking the ground is `v _(2),` then `v _(2) = v _(1) - (10 % of v _(1))` `= 42 -4.2` `= 37.8 ms ^(-1)` Height achieved by ball after striking ground is `h_(1),` then `h _(1) = (v _(2) ^(2))/( 2g) = (37.8 xx 37.8)/(2 xx 9.8) = 72.9m` Now when ball again strikes the ground, velocity is `V_(3),` then its displacemtn will be O. So, `v_(3) ^(2) - v _(2) ^(2) = 2g xx 0 [ because h =0]` `therefore v _(3) ^(2) = v _(2) ^(2)` `v _(3) = v _(2) = 37.8 ms ^(-1)` If time taken between two successive colisions is `t _(2),` then `0= v _(2) t _(2) + 1/2 (-g) t_(2) ^(2)` `0= 37.8 t _(2) - (9.8)/(2) t _(2) ^(2)` `4.9 t _(2) ^(2) = 37.8 t _(2)` `t _(2) = 7.714 s.` If velocity after second collision is `v _(4),` then `v _(4) = v _(3) - 10% of v _(3)` `= 37.8 -3.78` `v _(4) = 34.02 ms ^(-1)` Now `t _(2) = 7.714s` is time of flight (2T) `[because` Time taken to reach maximum height] ={Time taken to return back} So that `t _(2) = 2t.` `t. = (t_(2))/(2)` ` t . = ( 7.714)/(2) =3.857=3.86s` Now `t_(1) + t _(2) =4.28 + 7.714=11.994 s=12s`  Thus, ball will return back in `t =0` to t =12 s after colliding ground. Velocity of ball is `37.8 ms ^(-1)` when strikes first time on ground. |
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