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`        A ball dropped from the top of tower falls first half height of tower in 10 seconds . The total time spend by ball in air is`
4 years ago

```							s= ut + ½ at2Let the height of tower be 2h. h= 0 + ½ * g * t2=1/2 * g  * 102=50g 2h = ½ * g * T2100g = ½ * g * T2200 = T2T = 2001/2T = 10 root 2 seconds
```
4 years ago
```							Given that at t= 10 sec , distance travelled =h / 2Using the equation S=ut + (1/2)at2u =0 , a =10 m/s2 , t= 10 secS = 0 + (1/2) × 10×(10)2S = 500 mSo half the height of tower = 500 mi.e h/ 2 = 500 mh = 500×2 mh=1000 mTotal time taken by the ball can be determined by using the equation, S=ut + (1/2)at2Here S =1000 m , t = ?1000 = (1/2)×10×t2t2 = 2000 / 10t = 14.14 s
```
2 years ago
```							We have in the question the relationship b/w height and time .So, we keen to know a formula which gives relationship b/w time and heightA/c to motion equation , S=UT+1/2 aT^².Let us assume that total height be h.So,h/2 =ut +1/2 at^²
```
one year ago
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