Question

Two towers $A B$ and $C D$ are situated at distance $d$ apart as shown in figure. $A B$ is $20 m$ high and $C D$ is $30 m$ high from the ground. A particle is thrown from the top of $A B$ horizontally with a velocity of $10 {m s}^{- 1}$ towards $C D$ . Simultaneously, another particle is thrown from the top of $C D$ at an angle $60 °$ to the horizontal towards $A B$ with the same magnitude of initial velocity as that of the first object. The two particles moving in the same vertical plane collide in mid-air. Calculate the distance $d$ between the towers.

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Accepted Answer

Both the particles move under gravity, hence the relative acceleration is zero and consequently, the relative velocity between the particles will remain constant.

For two particles to collide, their relative velocity must be directed along the line joining them.

The relative velocity of the particle $C$ with respect to $d$ ,
${\vec{v}}_{C A} = {\vec{v}}_{C} - {\vec{v}}_{A} = {\vec{v}}_{C} + (- {\vec{v}}_{A})$

In given figure ${\vec{v}}_{C A}$ represents the relative velocity of the particle $C$ with respect to $A$ .

As $|{\vec{v}}_{C}| = |{\vec{v}}_{A}|$ , the resultant velocity would bisect the angle between ${\vec{v}}_{C}$ and $- v_{A}$ .

Hence, the angle between ${\vec{v}}_{C}$ and ${\vec{v}}_{C A}$ should be $30 °$ .

Thus $∠ C A E = 30 °$ .

Now in $Δ C A E$ ,

$\tan (30 °) = \frac{10}{d} \Rightarrow d = \frac{10}{\tan (30 °)} = 10 \sqrt{3} m$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 70: ILLUSTRATION 5.66

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 70: ILLUSTRATION 5.66

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