1. The problem statement, all variables and given/known data 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? 2. Relevant equations Vf^{2}=vi^{2} + 2ad 3. 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)= Vi^{2} + 2(-9.8)(1.652) -Vi^{2}= -32.38 Vi= 5.69 m/s This answer is incorrect? I dont understand how to do this??
In the problem the velocity of the player when he leaves the ground is required. So d is equal to Δd2.