Let, time taken is t.
So,
[tex]s=ut + \dfrac{gt^2}{2}\\\\75 = 0 + \dfrac{10\times t^2}{2}\\\\150=10\times t^2\\\\t = \sqrt{15}\\\\t = 3.873\ s[/tex] ( Here, us is vertical component of initial velocity )
Now, distance covered from the base of the cliff is :
[tex]D=5\times 3.873\ m\\\\D = 19.365\ m[/tex]
Therefore, time taken and distance covered is 3.873 s and 19.365 m.
Hence, this is the required solution.
