If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

If the equation of motion of a projectile is given by $y = 12x - \frac{3}{4}x^2$ and its horizontal component of velocity is $3 \ m/s$,then find its range. $(g = 10 \ m/s^2)$ (in $m$)

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle $45^o$ with the horizontal. The magnitude of the torque due to the weight of the projectile,when the particle is at its maximum height,about the point of projection is:

Two identical bodies,projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$,respectively,then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

$A$ $1.5 \,kg$ ball is shot upward at an angle of $34^{\circ}$ to the horizontal with an initial speed of $20 \,m/s$. The maximum height reached by the ball is (use $\cos 34^{\circ} = 0.83$ and $\sin 34^{\circ} = 0.56$): (in $\,m$)