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stepan [7]
3 years ago
15

A cannonball is fired horizontally from the top of a cliff. the cannon is at height h = 80.0 m above ground level, and the ball

is fired with initial horizontal speed v0. assume acceleration due to gravity to be g = 9.80 m/s2 .assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. what is the y position of the cannonball at the time tg/2?
Physics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer;

= 60 m

Explanation;

s = 80 m

= ½at² = ½ * 9.8m/s² * t² → t = 4.04 s  

At any time 0 ≤ t ≤ 4.04 s, height is  

y = Yo + ½at² = 80 m - 4.9m/s² * t²  

so at t = 4.04s/2 = 2.02 s,  

y = 80 m - 4.9m/s² * (2.02s)² = 60 m  

(so it drops 15 m during the first 2.02 s and 60 m during the second 2.02 s)

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
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Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

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    => a =  0.7156 m/s^2

     

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