A heavy particle is projected from a point on the horizontal at an angle $60^o$ with the horizontal with a speed of $10\ m/s$ . Then the radius of the curvature of its path at the instant of crossing the same horizontal will be ......... $m$
$20$
$30$
$25$
$30$
The velocity of projection of a body is increased by $2 \% .$ Other factors remaining unchanged, what will be the percentage change in the maximum height attained ? (in $\%$)
Given 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 same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$
Reason R: Product of said heights.
$h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)$
Choose the $CORRECT$ answer
Column $-I$ Angle of projection |
Column $-II$ |
$A.$ $\theta \, = \,{45^o}$ | $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$ |
$B.$ $\theta \, = \,{60^o}$ | $2.$ $\frac{{g{T^2}}}{R} = 8$ |
$C.$ $\theta \, = \,{30^o}$ | $3.$ $\frac{R}{H} = 4\sqrt 3 $ |
$D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$ | $4.$ $\frac{R}{H} = 4$ |
$K_i :$ initial kinetic energy
$K_h :$ kinetic energy at the highest point
A projectile is thrown with a velocity of $10\,m / s$ at an angle of $60^{\circ}$ with horizontal. The interval between the moments when speed is $\sqrt{5 g}\,m / s$ is $..........\,s$ $\left(g=10\,m / s ^2\right)$.
A particle is projected from a point $A$ with velocity $u\sqrt 2$ at an angle of $45^o$ with horizontal as shown in fig. It strikes the plane $BC$ at right angles. The velocity of the particle at the time of collision is