Physics Basketball player jumping problem?

Question

A 100 kg basketball player can leap straight up in the air to a height of 80 cm. You can understand how by analyzing the situation as follows: The player starts solid on the ground, with his knees bend ready to jump. THe instant he leaves the ground his head has moved vertically 60 cm, but remember his knees were bent to begin with. The max height from his feet to the ground is 80 cm
With what speed must the player leave the ground to reach a height of 80 cm?
m/s

What was his acceleration, assumed to be constant, as he jumped?

Answer 1

Some of this information is unnecessary, I'm pretty sure.
Using the equation of motion:
vf^2 = vi^2 + 2as
where vf is the final velocity, which is zero at the "top" of the jump.
vi is the unknown
a is the acceleration of gravity, which is -9.8 m/s/s
s is the distance, 80 cm or 0.8 m.

So, vi = sqrt(2*9.8*0.8)
this is about 4 meters/second

Answer 2

I'm making a few assumptions about this problem. They may or may not be accurate. 1) We're looking for maximum height. 2) The basketball player will attain maximum height in 0.689 s. 3) He starts from the ground. At maximum height, the velocity is zero: v2 = v1 + at v1 = v2 - at = 0 m/s - (-9.8 m/s^2)(0.689 s) = 6.7522 m/s After finding initial velocity, solve for height: x2 = (1/2)at^2 + v1t + x1 x2 = (1/2)(-9.8 m/s^2)(0.689 s)^2 + (6.7522 m/s)(0.689 s) + 0 m = 2.3261329 m He will attain about 2.3 m of vertical height. ==Edit: Oops, careless mistake. Forgot to square the time...heh heh.

Answer 3

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