Galileo writes that for angles of projection of a projectile at angles $(45 + \theta )$ and $(45 - \theta )$, the horizontal ranges described by the projectile are in the ratio of (if $\theta \le 45)$
$2:1$
$1:2$
$1:1$
$2:3$
Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.
$Column-I$ | $Column-II$ |
$(A)$ Angle of projection | $(p)$ $20\,m$ |
$(B)$ Angle of velocity with horizontal after $4\,s$ | $(q)$ $80\,m$ |
$(C)$ Maximum height | $(r)$ $45^{\circ}$ |
$(D)$ Horizontal range | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
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 $\%$)
A bullet is fired from a gun at the speed of $280\,ms ^{-1}$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is $........\,m$ $\left(g=9.8\,ms ^{-2}, \sin 30^{\circ}=0.5\right):-$
A body is projected with velocity $u$ making an angle $\alpha$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is
A ball is thrown at an angle $\theta $ and another ball is thrown at an angle $(90^o -\theta )$ with the horizontal from the same point with same speed $40\,ms^{-1}$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights?