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Two balls are thrown simultaneously from ground with same velocity of $10\,m / s$ but different angles of projection with horizontal. Both balls fall at same distance $5 \sqrt{3}\,m$ from point of projection. What is the time interval between balls striking the ground?
$(\sqrt{3}-1)\,s$
$(\sqrt{3}+1)\,s$
$\sqrt{3}\,s$
$1\,s$
Solution
(a)
$5 \sqrt{3}=\frac{(10)^2 \sin 2 \theta}{g} \text { or } \sin 2 \theta=\frac{\sqrt{3}}{2}$
$\therefore 2 \theta=60^{\circ} \text { or } \theta=30^{\circ}$
Two different angles of projection are therefore, $\theta$ and $\left(90^{\circ}-\theta\right)$ or $30^{\circ}$ and $60^{\circ}$.
Two different angles of projection are therefore, $\theta$ and $\left(90^{\circ}-\theta\right)$ or $30^{\circ}$ and $60^{\circ}$.
$T_1=\frac{2 u \sin 30^{\circ}}{g}=1\,s$
$T_2=2 \frac{u \sin 60^{\circ}}{g}=\sqrt{3}\,s$
$\Delta t=T_2-T_1=(\sqrt{3}-1)\,s$
Similar Questions
Match the columns
| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
