Newton's Laws of Motion

Question

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.280 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.860 m above the floor.

(a)

Calculate his velocity (in m/s) when he leaves the floor.

 __m/s

(b)

Calculate his acceleration (in m/s²) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.280 m. 

 __m/s²

(c)

Calculate the force (in N) he exerts on the floor to do this, given that his mass is 120 kg.

 __N

Answer 1

a) This part can be done with conservation of energy.

1. First, you can use your formula for Gravitational Potential Energy to calculate how much energy the player has at the top of the jump.
2. PE_g = mgh, where m is the mass of the player, g is the constant 9.81 m/s² and h is the height of the jump, 0.860 m. This will give you an answer in Joules.
3. Now, for the player to have this much energy at the top of this jump, they must've had it at the bottom too (since energy is conserved). With no other info given, we just assume that the gravitational potential energy all came from the kinetic energy that they had before they left the ground.
4. Just before they leave the ground, they have 0 J of gravitational potential energy. Therefore you can say that the answer to your first calculation equal to the kinetic energy just before leaving the ground
5. Simply put, PE_{g (top)} = KE_bottom
6. Recall that KE = 1/2 * m * v², and you should be able to solve for v.
7. SHORTCUT: Since both energy formulas have an "m" in them, once you get familiar with this type of problem, you can simplify to: gh = 1/2*v². Solving for v, this yields: v = √(2gh). CAREFUL: this will only work when the PE_g at the bottom is 0 and no other types of energy are involved.

b) You'll have to assume a constant acceleration here (always true in intro physics classes), and this is basically a 1-dimensional motion problem. There are a lot of formulas to get confused with here, and they get written a lot of ways, but you'll need to choose one that includes: a (because you're solving for it), d (because it's given), and v_final (which you just solved for). Also, do not forget that you do know the initial velocity (v_i) in this problem, because it's 0 m/s when the player's legs are fully bent.

1. I would use: v_f² = v_i² + 2ad.
2. It's not so bad when v_i = 0. Just plug in and solve for a, making sure to use the correct value for d, which is 0.280m, since that's the distance the player's mass moves while they are accelerating for their jump.

c) This is the easy part, where you just plug into F = ma, using your answer to part b and the mass given. Remember that forces are vectors, so you must give a direction. As it's asking for the force exerted by the player onto the floor, the direction for the force will be downwards.