Answer:
Part a)
![v_f = 7.99 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/9jm9tocmpfcrbut1twm8rojq28gv52eq51.png)
Part b)
![y = 3.25 m](https://img.qammunity.org/2020/formulas/physics/high-school/l3dricna9i7476v4wci82a977bnd85hbpa.png)
Step-by-step explanation:
Part a)
Since the diver is moving under gravity
so here its acceleration due to gravity will be uniform throughout the motion
so here we will have
![v_f^2 - v_i^2 = 2 a y](https://img.qammunity.org/2020/formulas/physics/high-school/s4b6di328ty7ozui1ouz73s8fx75trtwnc.png)
here we have
![v_i = 2.22 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/93n6ct04stcrl20yhj3jt3hqtt9ur2v1i2.png)
![y = -3 m](https://img.qammunity.org/2020/formulas/physics/high-school/ix6y3fg5iq9aktdj8xa1p8tqj9exsn3u2b.png)
![v_f^2 - (2.22)^2 = 2(-9.81)(-3)](https://img.qammunity.org/2020/formulas/physics/high-school/z38p3ikgvqs1key9tqxb7o4vms5qn1hezd.png)
![v_f = 7.99 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/9jm9tocmpfcrbut1twm8rojq28gv52eq51.png)
Part b)
at highest point of his motion the final speed will be zero
so we will have
![v_f^2 - v_i^2 = 2 a (\Delta y)](https://img.qammunity.org/2020/formulas/physics/high-school/n0lgwqv9xiyk4340l0x6emow9w3hl0ceni.png)
![0 - 2.22^2 = 2(-9.81)(y - 3)](https://img.qammunity.org/2020/formulas/physics/high-school/spcfbclqnrw41e9m39r6nmhz1z6hrgg6g1.png)
![y = 3.25 m](https://img.qammunity.org/2020/formulas/physics/high-school/l3dricna9i7476v4wci82a977bnd85hbpa.png)