Two spherical balls having equal masses with radius of \(5\, {cm}\) each are thrown upwards along the same vertical direction at an interval of \(3\, {s}\) with the same initial velocity of \(35 \,{m} / {s}\), then these balls collide at a height of \(\ldots \ldots \ldots . {m}\). (Take \({g}=10 \,{m} / {s}^{2}\) )

\({y}_{1}={y}_{2}\)

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

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

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

\(0=150-30 {t}\)

\({t}=5 {sec}\)

\(\therefore\) Height at which both balls will collied

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

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

\({h}=50 {m}\)