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`        wo balls A and B are thrown simultaneously, A vertically upwards with a speed of 20m/s from the ground, and B vertically downwards from a height of 100 m with the same seed and along the same line of motion. At what points do the two balls collide?`
5 months ago

```							Case 1:For the ball A :initialSpeed=u=20m/s[upwards]=-20m/sa=10m/s²[downwards]=+10m/s²time=t=tsecDistance=Sa metersFrom seond equation of motion,Sa=ut+1/2at²Sa=-20t+1/2x10xt²Sa=-20t+5t² --------------equation(1)Case II:For ball B:Initial speed=u=20m/s[downwards]=+20m/sa=10m/s²[downwards]=+10m/s²time =t secDistance =Sb meters.=(100 – Sa)From seond equation of motion,Sb=ut+1/2at²(4-Sa)=20t+5t² ====[equation2]Therefore , from equation 1 and 2 we get:40=-20t+5t²+20t+5t²40=10t²4=t²t=2secLet us find out distance Sa in 2 secSa=ut+1/2at²sa=-40+20=-20mFrom the ground that is at 20m the two balls will collide.
```
5 months ago
```							U1=20 m/s (upward)U2=20(downward)Let X be the distance from the ground where the collision takes place thenDistance from top of tower up to the collision point =40-XIf we suppose that collision took place after time T ThenX=20T-1/2×g×T^2……(1)(upward motion)40-X=20T+1/2g×T^2….(2)(downward motion)Adding (1) and (2) we get40=40TWhich gives T=1 secondPutting value of T in (1) we getX=20–1/2 ×9.8×1 ( taking :g= 9.8 m /s^2)X=20–4.9=15.1 metreSo collision takes at 15.1 metres from ground or 40–15.1=24.9 metres from top.
```
5 months ago
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