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.

 

[kerimek]

I would first clarify the units and values provided in the question. The height of 98.2 cm should be converted to meters (0.982 m) and the weight of the player should be converted to mass (98.6 kg).

Next, I would use the principle of conservation of energy to solve this problem. At the beginning of the jump, the player has potential energy due to his position and kinetic energy due to his speed. At the maximum height, the player has only potential energy. Therefore, we can equate the two energies to find the initial velocity needed for the player to reach a height of 0.982 m.

Using the equation for potential energy (PE=mgh) and setting it equal to the initial kinetic energy (KE=1/2mv^2), we get:

mgh=1/2mv^2

Rearranging for v, we get:

v=√(2gh)

Substituting in the values, we get:

v=√(2*9.8*0.982)

v=4.42 m/s

Therefore, the player must leave the ground with a speed of 4.42 m/s to reach a height of 0.982 m. This solution assumes that there is no air resistance and that all the energy is conserved throughout the jump. In reality, the player may need to leave the ground with a slightly higher speed to account for energy losses due to air resistance and friction.