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 |
$( A \rightarrow q , B \rightarrow q , r , C \rightarrow r , s , D \rightarrow s )$
$( A \rightarrow s , B \rightarrow q , r , C \rightarrow r , s , D \rightarrow p )$
$( A \rightarrow p , B \rightarrow q , r , C \rightarrow r , s , D \rightarrow s )$
$( A \rightarrow q , B \rightarrow q , p , C \rightarrow r , s , D \rightarrow s )$
A football player throws a ball with a velocity of $50$ metre/sec at an angle $30 $ degrees from the horizontal. The ball remains in the air for ...... $\sec$ $(g = 10\,m/{s^2})$
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.
The equation of motion of a projectile is $y=12 x-\frac{3}{4} x^2$ $..........\,m$ is the range of the projectile.
A ball thrown by one player reaches the other in $2\, sec$. The maximum height attained by the ball above the point of projection will be about .......... $m$
A body of mass $1\,kg$ is projected with velocity $50\,m / s$ at an angle of $30^{\circ}$ with the horizontal. At the highest point of its path a force $10\,N$ starts acting on body for $5\,s$ vertically upward besides gravitational force, what is horizontal range of the body? $\left(g=10\,m/s^2\right)$