A projectile is thrown with a velocity of $50\,\, ms^{^{-1}}$ at an angle of $53^o$ with the horizontal The equation of the trajectory is given by
$180y = 240x - x^2$
$180y = x^2 - 240x$
$180y = 135x - x^2$
$180y = x^2 - 135x$
A ball is thrown from a roof top at an angle of $45^o$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have $(a)$ greatest speed $(b)$ smallest speed $(c)$ greatest acceleration - Explain.
A particle is projected from ground with velocity $u$ at angle $\theta$ from horizontal. Match the following two columns.
Column $I$ | Column $II$ |
$(A)$ Average velocity between initial and final points | $(p)$ $u \sin \theta$ |
$(B)$ Change in velocity between initial and final points | $(q)$ $u \cos \theta$ |
$(C)$ Change in velocity between initial and final points | $(r)$ Zero |
$(D)$ Average velocity between initial and highest points | $(s)$ None of the above |
A particle is projected from the ground with velocity $u$ at angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are $R, H$ and $T$ respectively. They are given by $R = \frac{{{u^2}\sin 2\theta }}{g}$, $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$ Now keeping $u $ as fixed, $\theta$ is varied from $30^o$ to $60^o$. Then,
A projectile is launched at an angle ' $\alpha$ ' with the horizontal with a velocity $20 \; ms ^{-1}$. After $10 s$, its inclination with horizontal is ' $\beta$ '. The value of $\tan \beta$ will be : $\left( g =10 \; ms ^{-2}\right)$
A body is thrown with a velocity of $9.8 \,m/s$ making an angle of $30^o$ with the horizontal. It will hit the ground after a time ....... $\sec$