A ball is thrown from a point with a speed ${v_o}$ at an angle of projection $\theta $. From the same point and at the same instant a person starts running with a constant speed ${v_o}/2$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection
Yes, ${60^o}$
Yes, ${30^o}$
No
Yes, ${45^o}$
The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5{t^2})$ meter and $x = 6t\, meter$, where $t$ is in second. The velocity with which the projectile is projected is ......... $m/sec$.
The velocity of projectile at the intial point $A$ is $\left( {2\hat i + 3\hat j} \right)$ $m/s $ . It's velocity (in $m/s$) at point $B$ is
Two projectiles $A$ and $B$ are thrown with the same speed but angles are $40^{\circ}$ and $50^{\circ}$ with the horizontal. Then
A ball projected from ground at an angle of $45^o$ just clears a wall in front. If point of projection is $4\,m$ from the foot of wall and ball strikes the ground at a distance of $6\,m$ on the other side of the wall, the height of the walI is ........ $ m$
Two particles in same vertical plane are thrown to strike at same time. One from ground and other from height $h$ vertically above it. Ground particle is thrown obliquly and it achives a maximum height $H$. The second particle is thrown horizontally with same speed. What can be maximum $h$ so that two particles strike in air.