A body is thrown from a tower it covers 40m in last 2 second find height of tower
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Let h be the height of the tower and t the time taken\xa0u=0 and g= 10m/s2Using S= ut +1/2 gt2h= 0+5t2 —–(1)distance travelled in last 2 seconds is 40 magain using S= ut +1/2 gt2h-40= 0+5(t-2)2\xa0—–(2)subracting 2 from 1h-h+40= 5(t2\xa0- (t-2)28=\xa0(t2\xa0- (t-2)2\xa08= 4t-4t=3substituting t in equation 1h= 5x 3 x 3= 45 mHeight of tower = 45 m\xa0