How Fast Must a Basketball Player Jump to Achieve a 98.2 cm Vertical Leap?

[grewas8]

Homework Statement

A 98.6 kg basketball player can leap straight up in the air to a height of 98.2 cm, as shown below. The player bends his legs until the upper part of his body is dropped by 67.0 cm, then he begins his jump. With what speed must the player leave the ground to reach a height of 98.2 cm?
http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0848.png

Homework Equations

Vf²=vi² + 2ad

The Attempt at a Solution

because the jumper bends his legs before jumping, the delta d would be 98.2cm + 67.0cm= 1.652 m.
and because he reaches max height when velocity is zero,
(0)= Vi² + 2(-9.8)(1.652)
-Vi²= -32.38
Vi= 5.69 m/s

This answer is incorrect? I don't understand how to do this??

 

[rl.bhat]

In the problem the velocity of the player when he leaves the ground is required. So d is equal to Δd2.