Question

( 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} )

# 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)

Solution