The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$
$90$
$60$
$45$
$30$
A stone is projected from a point on the ground so as to hit a bird on the top of a vertical pole of height $h$ and then attain a maximum height $2 h$ above the ground. If at the instant of projection the bird flies away horizontally with a uniform speed and if the stone hits the bird while descending, then the ratio of the speed of the bird to the horizontal speed of the stone is
A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle $30^o$ with the horizontal. ........ $m$ far from the throwing point will the ball be at the height of $10\, m$ from the ground . $(g \,= \,10 m/s^2, \,sin \,30^o \,= \,\frac{1}{2}$, $\cos \,{30^o}\, = \,\frac{{\sqrt 3 }}{2}$)
A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3 s$ and $5 s$ after projection, during its flight. The speed of projection of the projectile is $.........\,ms ^{-1}$(Given $g=10\,m s ^{-2}$ )
A cricket fielder can throw the cricket ball with a speed $v_{0} .$ If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal, find
$(a)$ the effective angle to the horizontal at which the ball is projected in air as seen by a spectator
$(b)$ what will be time of flight?
$(c)$ what is the distance (horizontal range) from the point of projection at which the ball will land ?
$(d)$ find $\theta$ at which he should throw the ball that would maximise the horizontal range as found in $(iii)$.
$(e)$ how does $\theta $ for maximum range change if $u > u_0$. $u =u_0$ $u < v_0$ ?
$(f)$ how does $\theta $ in $(v)$ compare with that for $u=0$ $($ i.e., $45^{o})$ ?
The trajectory of a projectile near the surface of the earth is given as$ y = 2x -9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $(g = 10\, ms^{-2}$)