When a player throws a ball,it reaches the other player in $4 \,s$. If the height of each player is $1.8 \,m$,the maximum height attained by the ball above the ground is (in $\,m$)

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.

$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \, m$. With the same effort,he throws the ball vertically upwards. The maximum height attained by the ball is ......... $m$.

Two objects are projected with same velocity '$u$' at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^{\circ}$, the ratio of horizontal range of the first object to the second object will be $[\sin(\pi - \theta) = \sin \theta]$

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
Choose the $CORRECT$ answer.