If a ball projected vertically upwards with a certain initial velocity from the ground crosses a point at a height of $25 \ m$ twice in a time interval of $4 \ s$,then the initial velocity of the ball is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

The initial and final velocities of a body projected vertically from the ground are $20 \,ms^{-1}$ and $18 \,ms^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

From the top of a tower,a particle is thrown vertically downwards with a velocity of $10\; m/s$. The ratio of the distances covered by it in the $3^{rd}$ and $2^{nd}$ seconds of the motion is (Take $g = 10\; m/s^2$).

From the top of a tower,a ball is thrown vertically upward which reaches the ground in $6 \, s$. $A$ second ball thrown vertically downward from the same position with the same speed reaches the ground in $1.5 \, s$. $A$ third ball released from rest from the same location will reach the ground in ........ $s$.

When a ball is thrown up vertically with velocity $V_o$,it reaches a maximum height of $h$. If one wishes to triple the maximum height,then the ball should be thrown with what velocity?

$A$ 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: $(g = 10 \ m/s^2)$