A stone is dropped from the top of a tall tower and after one second another stone is dropped from a balcony 20m below the top. If both stones reach the ground at the same instant, calculate the height of the tower. (g =10m/s)

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( 20 int_{m}^{1} x^{1} frac{y^{prime}}{x^{2}} quad t_{1}=t_{1}=g, 8_{1}=(x+20) m ) ( x_{m} ) ( 8_{1}=mu_{1} t_{1}+frac{1}{2} a_{1} t_{1}^{2} ) ( x+20=frac{1}{2}(10) t_{1}^{2}- ) ( s_{2}=u_{2} t_{2}+frac{1}{2} a_{2} t_{2}^{2} ) ( x=frac{1}{2}(10)left(t_{1}-1right)^{2}- ) From (1) 8 2 ( frac{1}{2}(10)left(t_{1}-1right)^{2}+20=frac{1}{2}(10) t_{1}^{2} ) ( 5left(t_{1}^{2}=2 t_{1}+1right)+20=5 t_{1}^{2} ) ( 5 t_{1}^{2}-10 t_{1}+5+20=5 t_{1}^{2} ) ( t_{1}=2 cdot 5 s ) ( therefore x=5(2 cdot 5-1)^{2} ) ( x=5(1 cdot 5)^{2}=5 times 2 cdot 25=11 cdot 25 mathrm{m} ) T w ( begin{aligned} text { Height of tower } &=frac{11 cdot 25+2 Q}{[31-25 m} end{aligned} )

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