Two towers AB and CD are situated at distance d apart as shown in figure. AB is 20 m high and CD is 30 m high from the ground. A particle is thrown from the top of AB horizontally with a velocity of 10 ms^-1 towards CD. Simultaneously, another particle is thrown from the top of CD at an angle 60^o to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two particles move in the same vertical plane, collide in mid-air, calculate the distance d (in meter) between the towers. ${Take g=10ms}^{- 2}$

Moderate

Solution

Both the particles move under gravity, hence 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 A,
${\vec{v}}_{C, A} = {\vec{v}}_{C} - {\vec{v}}_{A} = {\vec{v}}_{C} + (- {\vec{v}}_{A})$
In given figure ${\vec{v}}_{C, A}$ represents the relative velocity of the particle C with respect to A. As $|{\vec{v}}_{C}| = |{\vec{v}}_{A}|,$ hence the angle between ${\vec{v}}_{C} and {\vec{v}}_{C, A}$ should be 30^o. Hence $∠ CAE = 30^{\circ}$ . Now in $ΔCAE$ ,
$\tan 30^{\circ} = \frac{10}{d} \Rightarrow d = \frac{10}{\tan 30^{\circ}} = 10 \sqrt{3} m = 17 .32 m$

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