Answer:
Explanation:
The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.
We will use the formula,
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]
[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]
Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.
Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]
