A swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the edge at the bottom ....... $m/s$ is $2\ m$ wide and $10\ m$ belows the top of the cliff .
$1.1$
$1.4$
$1.7$
$2.0$
The velocity of a body at time $ t = 0$ is $10\sqrt 2 \,m/s$ in the north-east direction and it is moving with an acceleration of $ 2 \,m/s^{2}$ directed towards the south. The magnitude and direction of the velocity of the body after $5\, sec$ will be
$Assertion$ : If a body is thrown upwards, the distance covered by it in the last second of upward motion is about $5\, m$ irrespective of its initial speed
$Reason$ : The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped.
The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) m$ at $t=$ $2 s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g =10\,ms ^{-2}$ )
Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are $v_1$ and $v_2$ at angles $\theta _1$ and $\theta_2$ respectively from the horizontal, then answer the following question
The trajectory of particle $1$ with respect to particle $2$ will be