${y}_{1}={y}_{2}$

$35 {t}-\frac{1}{2} \times 10 \times {t}^{2}=35({t}-3)-\frac{1}{2} \times 10 \times({t}-3)^{2}$

$35 {t}-\frac{1}{2} \times 10 \times {t}^{2}=35 {t}-105-\frac{1}{2} \times 10 \times {t}^{2}$

$\quad-\frac{1}{2} \times 10 \times 3^{2}+\frac{1}{2} \times 10 \times 6 {t}$

$0=150-30 {t}$

${t}=5 {sec}$

$\therefore$ Height at which both balls will collied

${h}=35 {t}-\frac{1}{2} \times 10 \times {t}^{2}$

$=35 \times 5-\frac{1}{2} \times 10 \times 5^{2}$

${h}=50 {m}$