Ask question

Ask question

All categories

3 years ago

5

You and your best friend are bored on a saturday afternoon and decide to measure the impulse acting on a basketball when it boun

ces off the ground. the ball with a mass of m is dropped from rest at a height h. assume no fritcional forces are in play such that the ball bounces as a perfectly elastic colision.
derive an experession for the velocity of the ball jus before it collides with the ground.

2 answers:

grigory [225]3 years ago

7 0

Answer:

v = √(2gh)

Assuming g = 9.81m/s^2

v = √(19.62h)

Explanation:

Let v represent the speed of the ball just before it collide with the ground.

Applying the equation of motion;

v^2 = u^2 + 2as

Where;

v = final velocity

u = initial speed = 0 (starting from rest)

a = acceleration= g acceleration due to gravity

s = distance covered = h

So, substituting the values;

v^2 = 0^2 + 2gh

v^2 = 2gh

v = √(2gh)

Assuming g = 9.81m/s^2

v = √(19.62h)

Burka [1]3 years ago

4 0

Answer:

V = (19.62h)^0.5

Explanation:

We can apply the principle of conservation of momentum in this case.

The potential energy due to the height of fall is transformed into the kinetic energy of the ball as it falls.

PE = mgh

Where m is the mass of the ball,

g is the acceleration due to gravity = 9.81 m/s2,

h is the height of fall.

The kinetic energy of the ball = 1/2(mv^2)

Where c is the velocity of the ball.

Equating both energy,

mgh = 1/2(mv^2)

2gh = v^2

v = (2gh)^0.5 = (19.62h)^0.5

Or velocity V is equal to the square root of (19.62h)

You might be interested in

When you are ice skating, to get started, you push your stake backwards on the ice and, as a result, begin to move forward. whic

LUCKY_DIMON [66]

C. Newtons third law of motion

Because eventually, the frictional forces will slow you to a halt. Newton's Third Law of Motion For every action there is an equal and opposite reaction. When they push off against the ice, or "stroke" with their skates, they are applying a force down and back against the ground.

Hope this helps!

8 0

2 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago

Which best describes the current atomic theory?

ivolga24 [154]

Choice-C is a correct statement.

6 0

2 years ago

Read 2 more answers

What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

7 0

3 years ago

Is blacklight the same as ultraviolet?

Misha Larkins [42]

Blacklight is made of a type of ultraviolet light called UV-A, but blacklight is not the same as ultraviolet light. Blacklight lamps only emit in the lower energy range of ultraviolet light.

6 0

2 years ago