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 of 35 m/s, then these balls collide at a height of .......... m. (Take, g = 10 m/s²)

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answer is 50.

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Detailed Solution

$\begin{array}{l} Given, m_{1} = m_{2} = m \\ r_{1} = r_{2} = r = 5 cm = 5 \times 10^{- 2} m \\ u_{1} = u_{2} = 35 m / s \\ t_{1} = 0 \\ and t_{2} = 3 s \end{array}$

When both ball will collide,

${(S_{1})}_{vertical} = {(S_{2})}_{vertical} . . . (i)$

Let after time t both ball will collide.

$\begin{array}{l} {(S_{1})}_{vertical} = u_{1} t - \frac{1}{2} {gt}^{2} \\ {(S_{2})}_{vertical} = u_{2} (t - 3) - \frac{1}{2} g (t - 3)^{2} \end{array}$

From Eq. (i), we have

$\begin{array}{l} \Rightarrow u_{1} t - \frac{1}{2} {gt}^{2} = u_{2} (t - 3) - \frac{1}{2} g (t - 3)^{2} \\ \Rightarrow 35 t - \frac{1}{2} \times 10 \times t^{2} = 35 (t - 3) - \frac{1}{2} \times 10 (t - 3)^{2} \\ \Rightarrow 35 t - 35 (t - 3) = 5 t^{2} - 5 (t - 3)^{2} \\ \Rightarrow 35 (t - t + 3) = 5 [t^{2} - (t - 3)^{2}] \\ \Rightarrow 35 \times 3 = 5 (t + t - 3) (t - t + 3) \\ \Rightarrow \frac{105}{5} = (2 t - 3) (3) \\ \Rightarrow \frac{21}{3} = 2 t - 3 \\ \Rightarrow 7 = 2 t - 3 \\ \begin{array}{l} \Rightarrow 2 t = 10 \\ \Rightarrow t = 5 s \end{array} \end{array}$

So, the height at which ball will collide is

$\begin{array}{l} {(S_{1})}_{vertical} = u_{1} t - \frac{1}{2} {gt}^{2} \\ = 35 \times 5 - \frac{1}{2} \times 10 \times (5)^{2} \\ = 175 - 125 \\ = 50 m \end{array}$