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(B) Given that the height achieved in $1 \,s$ is $h=25 \,m$. Let $u$ be the initial velocity in the upward direction. Using the equation of motion, $h

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$\frac{4}{5}$

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(B) Given that the height achieved in $1 \,s$ is $h=25 \,m$. Let $u$ be the initial velocity in the upward direction. Using the equation of motion, $h=ut-\frac{1}{2}gt^2$.$25=u(1)-\frac{1}{2}(10)(1)^2$$25=u-5 \Rightarrow u=30 \,m/s$.The final velocity of the ball at maximum height becomes zero. The time taken to reach the maximum height is $v=u-gt \Rightarrow 0=30-10t \Rightarrow t=3 \,s$.Now, the distance travelled in $2 \,s$ is the displacement in $2 \,s$ since the ball is still moving upwards:$d_1=u(2)-\frac{1}{2}g(2)^2 = 30(2)-5(4) = 60-20 = 40 \,m$.The maximum height achieved by the ball in $3 \,s$ is:$H=u(3)-\frac{1}{2}g(3)^2 = 30(3)-5(9) = 90-45 = 45 \,m$.The displacement after $4 \,s$ is:$S_2=u(4)-\frac{1}{2}g(4)^2 = 30(4)-5(16) = 120-80 = 40 \,m$.Since the ball has passed the maximum height at $t=3 \,s$, the total distance travelled in $4 \,s$ is $d_2=H+(H-S_2) = 45+(45-40) = 50 \,m$.The ratio is $\frac{d_1}{d_2} = \frac{40}{50} = \frac{4}{5}$.

$A$ ball thrown vertically upwards reaches a height of $25 \,m$ in $1 \,s$. The ratio of the total distance covered by the ball in time $t=2 \,s$ and $t=4 \,s$ is (Use $g=10 \,m/s^2$)

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