The rock will be at 90.4 m from the top of the cliff.
Explanation:
The rock is thrown with the “initial velocity” 3 m/s. We need to find how much distance does the rock traveled in 4 seconds (t).
From the “kinematic equations” take
[tex]s=u t+\frac{1}{2} a t^{2}[/tex]
Where, “s” is distance traveled, “u” initial velocity of the object, “t” time the object traveled and “a” acceleration due to gravity is [tex]9.8 \mathrm{m} / \mathrm{s}^{2}.[/tex]
Substitute the given values in the above formula,
[tex]s=3 \times 4+\frac{1}{2} \times 9.8 \times 4^{2}[/tex]
[tex]s=12+\frac{1}{2} \times 9.8 \times 16[/tex]
[tex]s=12+\frac{1}{2} \times 156.8[/tex]
[tex]s=12+78.4[/tex]
[tex]s=90.4[/tex]
The rock is at height of 90.4 m from the top of the cliff.
