$A$ particle is projected horizontally from a tower with velocity $10\,m/s$. Taking $g=10\,m/s^2$,match the following two columns at time $t=1\,s$.
Column $I$ Column $II$ $(A)$ Horizontal component of velocity $(p)$ $5$ $SI$ unit $(B)$ Vertical component of velocity $(q)$ $10$ $SI$ unit $(C)$ Horizontal displacement $(r)$ $15$ $SI$ unit $(D)$ Vertical displacement $(s)$ $20$ $SI$ unit
| Column $I$ | Column $II$ |
| $(A)$ Horizontal component of velocity | $(p)$ $5$ $SI$ unit |
| $(B)$ Vertical component of velocity | $(q)$ $10$ $SI$ unit |
| $(C)$ Horizontal displacement | $(r)$ $15$ $SI$ unit |
| $(D)$ Vertical displacement | $(s)$ $20$ $SI$ unit |
- A$(A \rightarrow q, B \rightarrow q, C \rightarrow q, D \rightarrow p)$
- B$(A \rightarrow q, B \rightarrow r, C \rightarrow q, D \rightarrow p)$
- C$(A \rightarrow q, B \rightarrow s, C \rightarrow q, D \rightarrow p)$
- D$(A \rightarrow s, B \rightarrow q, C \rightarrow q, D \rightarrow p)$
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View SolutionGiven below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
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Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
Choose the $CORRECT$ answer.
$A$ particle is projected with speed $u$ at an angle $\theta$ with the horizontal from the ground. If it is at the same height from the ground at times $t_1$ and $t_2$,then its average velocity in the time interval $t_1$ to $t_2$ is .........
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View Solution$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be
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View Solution$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .
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