Question

A ball is dropped from bridge of 122.5 m above river. After the ball has been falling for 2 seconds ,a second ball is thrown straight down after it. Initial velocity of second ball ao that both hit the water at same time.
(1) 49m/sec
(2) 55.5m/sec
(3) 26.1m/sec
(4)9.8m/sec

Answer 1

Answer on Question #45187, Physics, Mechanics | Kinematics | Dynamics

A ball is dropped from bridge of $122.5\mathrm{m}$ above river. After the ball has been falling for 2 seconds, a second ball is thrown straight down after it. Initial velocity of second ball ao that both hit the water at same time.

(1) $49\mathrm{m / sec}$

(2) $55.5\mathrm{m / sec}$

(3) $26.1\mathrm{m / sec}$

(4) $9.8\mathrm{m / sec}$

Solution

Let $t_1$ be the time of flight of the first ball and $t_2$ for the second one.

Then

$$t _ {1} = t _ {2} + \Delta t \rightarrow t _ {2} = t _ {1} - \Delta t$$

Where $\Delta t = 2s$ . Equations of motion for the both balls:

$$h = \frac {g t _ {1} ^ {2}}{2} \rightarrow t _ {1} = \sqrt {\frac {2 h}{g}}$$ $$\begin{array}{l} h = V _ {0 2} t _ {2} + \frac {g t _ {2} ^ {2}}{2} = V _ {0 2} t _ {2} + \frac {g t _ {2} ^ {2}}{2} = V _ {0 2} (t _ {1} - \Delta t) + \frac {g (t _ {1} - \Delta t) ^ {2}}{2} \\ = V _ {0 2} \left(\sqrt {\frac {2 h}{g}} - \Delta t\right) + \frac {g \left(\sqrt {\frac {2 h}{g}} - \Delta t\right) ^ {2}}{2} \\ \end{array}$$

So:

$$V_{02} = \frac{h - \frac{g \left(\sqrt{\frac{2h}{g}} - \Delta t\right)^2}{2}}{\sqrt{\frac{2h}{g}} - \Delta t} = \frac{h}{\sqrt{\frac{2h}{g}} - \Delta t} - \frac{g \left(\sqrt{\frac{2h}{g}} - \Delta t\right)}{2}$$

Numerically:

$$V_{02} = \frac{122.5\,m}{\sqrt{\frac{2 \cdot 122.5\,m}{9.8\, \frac{m}{s^2}}} - 2s} - \frac{9.8\, \frac{m}{s^2} \cdot \left(\sqrt{\frac{2 \cdot 122.5\,m}{9.8\, \frac{m}{s^2}}} - 2s\right)}{2} \approx 26.1\, \frac{m}{s}$$

Answer: (3) 26.1m/sec

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