Kinematics - Grabbing a Rebound?

[johndoe3344]

A basketball player grabbing a rebound jumps 73.0 cm vertically. How much total time (ascent and descent, in ms) (a) does the player spend in the top 13.2 cm of this jump? (b) in the bottom 13.2 cm?

(a)

The formula I use is dx = Vot + 0.5at^2

Vo = 0 (at the top of the jump)
dx = -.132 m
a = -9.8 m/s^2

t = .164

I figure that the time spent going up and the time spent going down should be exactly the same.

So 2t = .328 seconds

(b)

I use the formula Vf^2 = Vo^2 + 2adx

Vf = 0 (at the top of the jump)
dx = 0.73 m
a = -9.8 m/s^2

Vo = 3.78 m/s

With Vo, I use the formula dx = Vot + 0.5at^2

dx = .132 m
a = -9.8 m/s^2

And I get t = .0366

So 2t = .0732 seconds

Is everything correct?

 

[Andrew Mason]

A basketball player grabbing a rebound jumps 73.0 cm vertically. How much total time (ascent and descent, in ms) (a) does the player spend in the top 13.2 cm of this jump? (b) in the bottom 13.2 cm?

(a)

The formula I use is dx = Vot + 0.5at^2

Vo = 0 (at the top of the jump)
dx = -.132 m
a = -9.8 m/s^2

t = .164

Not quite. You are finding the first 13.2 cm going up, so v0 is not 0. Draw a graph of the speed, v, vs. time t. The area under the graph is distance.

$$d = v_{avg}t = \frac{1}{2}(v_0 + v_{13.2})t$$

where:
$$v_{13.2} = v_0 - gt$$It is just the reverse for the last 13.2 cm. By symmetry you can see that the speed at 0 and 13.2 cm, respectively, going up is the same at these points going down.

So for the last 13.2 cm:

$$d = v_{avg}t = \frac{1}{2}(v_{13.2}+ v_0)t$$So the time spent in the first 13.2 cm should be exactly the same as the time spent in the last 13.2 cm.

AM

 

[johndoe3344]

Not quite. You are finding the first 13.2 cm going up, so v0 is not 0.

I was calculating the time of the upper 13.2 cm going down - since it's the same as the 13.2 cm going up and it's easier because v0 is 0. Does the problem not work that way? Thanks.

 

[Doc Al]

Is everything correct?

I didn't confirm your arithmetic, but your methods are exactly correct.

 

[Andrew Mason]

I was calculating the time of the upper 13.2 cm going down - since it's the same as the 13.2 cm going up and it's easier because v0 is 0. Does the problem not work that way? Thanks.

I was using v0 is the speed at the beginning of the jump. It is not any easier to use v0 as the speed at the top because you have to still figure out the speeds at the beginning and end of the interval, so you still have to work out speed at 13.2 cm and 0 (height). All you have to know is that the time spent over a particular distance is the distance covered divided by the average speed. Where acceleration is constant, that average is just one half of the sum of the speeds at the beginning and at the end of that interval.

AM

PS. Despite what Doc Al says, your general formula is correct but you are not applying it correctly. The d in your formula is the distance traveled from 0 (ie the point at which v = v0). You could use your formula to determine the time it takes to travel from the top to a point 13.2 cm above the ground. You could also use it to determine the time it takes to travel to the ground and then subtract the time it took to reach 13.2 cm.

 

[Doc Al]

(a)

The formula I use is dx = Vot + 0.5at^2

Vo = 0 (at the top of the jump)
dx = -.132 m
a = -9.8 m/s^2

t = .164

I figure that the time spent going up and the time spent going down should be exactly the same.

So 2t = .328 seconds

Your method, and answer, seem completely correct. There is nothing at all wrong with starting at the top and figuring out the time it takes to fall 0.132m.

(b)

I use the formula Vf^2 = Vo^2 + 2adx

Vf = 0 (at the top of the jump)
dx = 0.73 m
a = -9.8 m/s^2

Vo = 3.78 m/s

With Vo, I use the formula dx = Vot + 0.5at^2

dx = .132 m
a = -9.8 m/s^2

And I get t = .0366

So 2t = .0732 seconds

Again, your method is exactly right. Check the solution of your quadratic equation when solving for t. (I get a different answer.)

 

[Andrew Mason]

Sorry. I misread the question. Your method is quite correct. I was calculating the 13.2 cm as the distance at the bottom going up and coming down, not at the top. My apologies to you and Doc Al.

AM