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julsineya [31]
3 years ago
11

Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿

a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Physics
1 answer:
FrozenT [24]3 years ago
4 0

Answer:

Tenemos dos problemas a resolver acá:

Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.

Acá podemos usar la conservación de la energía.

E = U + K

U = energía potencial = m*g*H

m = masa

g = aceleración gravitatoria = 9.8m/s^2

H = altura

K = energía cinética = (m/2)*V^2

donde V es la velocidad.

Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:

Ei = U = m*(9.8m/s^2)*2.5m

Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:

Ef = (m/2)*V^2

Y como la energía se conserva, la energía final es igual a la inicial, entonces:

m*(9.8m/s^2)*2.5m = (m/2)*V^2

Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.

V = √(2*(9.8m/s^2)*2.5m) = 7m/s

Ahora respondamos la segunda parte.

Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:

A(t) = -9.8m/s^2

Para obtener su velocidad integramos:

V(t) = (-9.8m/s^2)*t + V0

donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s

V0 = (3/4)*7m/s = (21/4) m/s

Así, la ecuación de la velocidad es:

V(t) = (-9.8m/s^2)*t + (21/4) m/s

Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:

V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s

         t =  (21/4) m/s/9.8m/s^2 = 0.54 s

Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.

Para ello integramos de vuelta:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0

donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t  

La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s

P(0.54s) =  (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m

La altura máxima es 1.81 metros.

Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.

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Answer:

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Explanation:

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acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

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we substitute

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so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

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Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

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t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

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t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

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Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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