The equation of motion of a projectile is: $y = 12x - \frac{5}{9}{x^2}$. The horizontal component of velocity is $3\ ms^{- 1}$ . Given that $g = 10\ ms^{- 2}$ , .......... $m$ is the range of the projectile .
$12.4$
$21.6$
$30.6$
$36.0$
At the top of the trajectory of a projectile, the directions of its velocity and acceleration are
Given that $u_x=$ horizontal component of initial velocity of a projectile, $u_y=$ vertical component of initial velocity, $R=$ horizontal range, $T=$ time of flight and $H=$ maximum height of projectile. Now match the following two columns.
Column $I$ | Column $II$ |
$(A)$ $u_x$ is doubled, $u_y$ is halved | $(p)$ $H$ will remain unchanged |
$(B)$ $u_y$ is doubled $u_x$ is halved | $(q)$ $R$ will remain unchanged |
$(C)$ $u_x$ and $u_y$ both are doubled | $(r)$ $R$ will become four times |
$(D)$ Only $u_y$ is doubled | $(s)$ $H$ will become four times |
A ball of mass $m$ is thrown vertically upwards. Another ball of mass $2m$ is thrown at an angle $\theta $ with the vertical. Both of them stay in air for same period of time. The heights attained by the two balls are in the ratio of
A projectile is fired at $30^{\circ}$ to the horizontal, The vertical component of its velocity is $80 \;ms ^{-1}$, Its time flight is $T$. What will be the velocity of projectile at $t =\frac{ T }{2}$?
A projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$. At the same time another projectile $B$ is thrown with velocity $v_2$ upwards from the point $Q$ vertically below the highest point $A$ would reach. For $B$ to collide with $A$, the ratio $\frac{v_2}{v_1}$ should be