Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of35 m/s, then these balls collide at a height of $\_\_\_\_\_\_$ m.

From the figure when both balls collide

${\mathrm{y}}_1={\mathrm{y}}_2$

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

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

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

$$\begin{gathered} 0=150-30\mathrm{t} \\ \Rightarrow \mathrm{t}=5\sec \end{gathered}$$

Thus the height at which balls will collide is,

$\mathrm{h}=35\mathrm{t}-\frac{1}{2}\times 10\times {\mathrm{t}}^2=35\times 5-\frac{1}{2}\times 10\times 5^2=50\mathrm{m}$

Hence the correct answer is 50m.