$A$ ball is thrown vertically upwards with an initial velocity $u$ and reaches its maximum height in $5 \,s$. The ratio of the distance travelled by the ball in the $2^{nd}$ and $7^{th}$ second is (assume $g=10 \,m/s^2$):

$Assertion$ : Two balls of different masses are thrown vertically upward with the same speed. They will pass through their point of projection in the downward direction with the same speed.
$Reason$ : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball.

If the distance travelled by a freely falling body in the last but one second of its motion is $5 \ m$,then the total time of motion is (Acceleration due to gravity $g = 10 \ m \ s^{-2}$) (in $s$)

Two balls are released from the same position at a height of $500 \ m$ above the ground, one after the other, with an interval of $1 \ s$. What is the distance between the two balls when the first ball hits the ground (in $m$)? (Acceleration due to gravity $g = 10 \ m/s^2$)

$A$ ball is thrown vertically upwards with a velocity of $19.6 \, m/s$ from the top of a tower. The ball strikes the ground after $6 \, s$. The height from the ground up to which the ball can rise is given by $\left(\frac{k}{5}\right) \, m$. The value of $k$ is ..... (use $g = 9.8 \, m/s^2$)

$A$ ball is thrown vertically upwards with a speed of $10 \ m/s$ from the top of a tower $200 \ m$ high and another is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground (in $s$) is: $(g = 10 \ m/s^2)$