# Physics: Does anyone have an idea how long a juggling cube takes to be released.?

I am having trouble estimating the amount of time it takes from the time the hand begins its upward motion to where the cube leaves the hand and the hand stops its upward motion. I am getting a ridiculous answer like 1/20 of a second and I know that is not possible. I think the release distance is about 6 cm and the height the cube ultimately reaches is about 30 cm for a juggler who is just practising for exercise.

The cube weighs 90 grams if that is useful. If you need further information either email me or ask me to revise the question with additional details.

I always award 10 points if that means anything.

Relevance

Let us star by working out the velocity of the cube as it leaves the hand:

Equation:

V² = u² + 2*a*s

V = final velocity = 0m/s

u = initial velocity - what we want

a= acceleration due to gravity = -9.81m/s² ( negative because acceleration is downwards)

S = displacement = 30cm = 0.3m

0 = u² + 2*(-9.81)*0.3

u² = 5.886

u = 2.43m/s

You state that the hand moves 6cm in increasing the velocity of the cube from 0 to 2.43m/s

In order to calculate the time taken to do this we need the acceleration -

Again we use

v² = u² + 2as

2.43² = 0 + 2*a*0.06

5.886 = 0 + 0.12a

a = 5.886/0.12

a = 49.05m/s²

To calculate the time it takes from the time the hand begins its upward motion to where the cube leaves the hand:

v = u +at

2.43 = 0 + 49.05t

t = 2.43/49.05

t = 0.0495 seconds

If you call this 0.05 seconds - this is 1/20th second which does not seem to be so ridiculous.

• Anonymous
7 years ago

total distance moved by cube = 30 - 6 = 24cm = 0.24 m

acceleration due to gravity = g = 9.8 m/s^2

time going upward = time coming downward

now while coming downward ,

initial velocity of cube = u = 0

Newton's equation of motion,

s = ut + (1/2)*g*t^2

0.24 = 0 + (1/2)*9.8*t^2

solve for t.

u get ,

t = 0.22 sec.

which is quite practical because g is quite high to acquire velocity and distance is very small!

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