$A$ ball is thrown vertically upward. It has a speed of $10 \; m/s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take $g = 10 \; m/s^2$)

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A = 6 \ s$ and $t_B = 2 \ s$,then the velocity $u$ of each ball is ....... $m/s$.

$A$ projectile is fired vertically upwards with an initial velocity $u$. After an interval of $T$ seconds,a second projectile is fired vertically upwards,also with initial velocity $u$.

$A$ juggler throws balls vertically upwards with the same initial velocity in the air. When the first ball reaches its highest position,he throws the next ball. Assuming the juggler throws $n$ balls per second,the maximum height the balls can reach is

$A$ body is dropped from a height $H$. The time taken to cover the second half of the journey is:

$A$ stone is thrown upwards with a velocity $50 \,ms^{-1}$. Another stone is simultaneously thrown downwards from the same location with a velocity $50 \,ms^{-1}$. When the first stone is at the highest point, the velocity of the second stone is $...... \,ms^{-1}$ (Take $g = 10 \,ms^{-2}$).