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`        a body dropped from certain height travels 176.4 m in the last 2 seconds , then find the height from which the body was dropped `
10 years ago 8 Points
```							Let T and H be the total time of flight and height respectively.
Then H = 0.5 * g * T^2
Let v m/s be the velocity of the particle at the beginning of last  2 second.
v = g*(T-2)
according  to the problem, 176.4 = v*2 + 0.5 * g  * 2^2
= 4gT -2g
=> T = (176.4 + 2*9.8)/(4*9.8)
= 5
Hence H = 0.5 * 9.8 * 25 = 122.5
```
10 years ago
```							let u be the vel. before the 2 sec. then.........
s=176.4    a=9.9         t=2s
s=ut+1/2at2
176.4=u*2+1/2*9.8*4
u=78.4 m/s
this u will be the final velocity for the block when dropped from heighest point before last 2 sec....
then,
u=0  ,v=78.4   ,a=9.8,  s=?
2as=v*2-u*2
on calculating   s=313.6m
total h=313.6+176.4=490m

```
10 years ago
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