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3 years ago

```							Dear student,Please post one question per thread only,Answer to question number 15-Let the height of point from ground where they meet beand time after which they meet beHence for first ball,For the second ball,Adding the two gives, t=4s, h=20m
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2 years ago
```							Take height from the top as x When dropped from top , u = 0, g = 9.8 , h = x, time taken is tx=ut+0.5gt2 = 0 + 0.5(9.8)t2= 4.9t2   (1) When the  ball projected upward, h = 100-x , u = 25 m/s , time is same t100-x = 25 t-0.5*9.8*t2x = 100-25t+4.9t2 (2) from (1) and (2)4.9t2 = 100-25t+4.9t2t = 100/25 = 4 seconds From (1)x= 4.9(4)2    = 78.4 meters
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2 years ago
```							For 14Let height reached by particle on the top of building be xInitial velocity u= 14m/sAnd final velocity v =0As v= u+atv= u- gtt= 1.42sAs h= ut -1/2gt2 h = 14×1.42- 1/2 ×10×1.42×1.42   = 19.88- 10.08    = 9.8mGiven total time taken= 4.6sTime left 4.6-1.42= 3.18sSecond case::Velocity at heighest point is zerou=0v =? t=3.18sv= 9.8×3.18= 31.164m/s2as= v2 - u2s= (31.164×31.164)/2×9.8  = 49.55mTotal height = 49.55mHeight of building = 49.55-9.8  = 39.7m
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2 years ago
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