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`        A ball is thrown vertically upward from the top of a tower with an initial velocity of 19.6 metre per second the ball which is the ground after 5 seconds. calculate first the height of the tower and the velocity of ball reaching the ground take g is equals to 9.8 metre per second square`
3 years ago

## Answers : (1)

Vikas TU
10277 Points
```							From the top of the tower when the ball is thrown and when it reaches the bottom again,the Time taken total = 2u/g => 4 seconds that means it takes 5-4 =1 s to travel the tower height.and max. height it takes = > u^2/2g => 19.6*19.6/2*9.8 => 19.6 meter.From first law of eqn. we get,v = 19.6 – g*4 => -19.6 m/s.Hence from second’s law of eqn. we get,h = 19.6*1 + 0.5*9.8*1^1 = > 24.5 meter.
```
3 years ago
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