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2 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]2 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]2 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)

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tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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2 years ago

What is one benefit of lifelong physical activity?

kvasek [131]

The answer would be C

5 0

2 years ago

A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because

amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

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2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

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2 years ago

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mr_godi [17]

Answer:

D, I think.

Explanation:

I had a quiz in Plate Tectonics and there was 2 questions that are related to this, but not the exact question.

Which material rises from cracks in oceanic crust

-molten rock

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those are all right btw, so you can decide if the answer I told you is right or not.

4 0

1 year ago

Read 2 more answers