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

$\mathrm{s}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2\Rightarrow 25=\mathrm{u}-5\Rightarrow \mathrm{u}=30\mathrm{m}/\mathrm{s}$

Time of ascent = 3s.

At t = 3s the ball comes to rest and changes the direction of motion.

Distance travelled in t=2 is given by ${\mathrm{s}}_2=30\left(2\right)-\frac{1}{2}10{\left(2\right)}^2=40\mathrm{m}$

Maximum height = $\frac{{\mathrm{u}}^2}{2\mathrm{g}}=45\mathrm{m}$, and from t=3 to t=4 it traveled through a distance of $\frac{1}{2}10{\left(1\right)}^2=5\mathrm{m}$

So the ratio of distance = 4/5