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Lesechka [4]
3 years ago
9

Three balls are thrown off the top of a building, all with the same speed but with different launch angles (position is given in

meters and time is given in seconds). The components of the initial velocities are given.
The blue ball has an initial velocity of (6 m/s, 8 m/s).
The green ball has an initial velocity of (10 m/s, 0 m/s).
The red ball has an initial velocity of (8 m/s, -6 m/s).

a. Rank the three balls according to which one hits the ground first.
b. Rank the three balls according to which one has the greatest speed the instant before impact with the ground.
c. Calculate the speed of each of the balls the instant before impact with the ground.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

A. Red, Green, Blue

B. None. They all hit the ground with the same speed.

C. Final Kinetic Energy = Initial Gravitational Potential Energy

Explanation:

A. This one is actually just common sense. If you think about it, the object that is being thrown downward will reach the ground first. The object thrown upwards will hit the ground the last. If you don't believe me, use just the vertical component of velocity and use acceleration = 9.8 m/s^2. Find the time it takes for an arbitrary displacement each ball will travel. You will find that the one pointed downward will reach the ground in the shortest amount of time. Just do the same math over for the rest of the balls and you can find the time it takes for each ball to hit the ground.

B. This one is actually really simple if you think about it. Using the Work-Energy theorem, using the system consisting a ball and the earth there will be no external forces doing work on the system. Since each ball is at some height "h" and the same initial speed, they will all have the same amount of energy in the beginning. If we set the ground as our reference point for gravitational potential energy, then there will be no gravitational energy as the height at that point will be 0. They all must have the same kinetic energy at the ground.

C.

Mechanical Energy Final = Mechanical Energy Initial + Net Work from non-conservative forces

Kinetic Energy Final + Gravitational Potential Energy Final = Kinetic Energy Initial + Gravitational Potential Energy Initial

I am using the symbol " ' " to indicate final speeds and heights

0.5*m*v'^2 + mgh' = 0.5*m*v^2 + mgh

0.5*m*v'^2 + 0 = 0.5*m*v^2 + mgh

(cancel out the mass)

0.5*v'^2 = 0.5*v^2 +gh

Since the "v" is the same initially for all three balls and "g" and "h" are the same for all three balls. The final speed v' must be the same for all three situations regardless of mass.

If you have any questions feel free to comment again. I'll try to clarify.

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5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.
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Answer:

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
  • Let's draw the free body diagram of the system.
  • To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

                      F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\

  • To find the answer, we have to find the tension,

                     Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N

  • Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

                F_V=(29*9.8)-(169.43*sin57)=142.10N

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

5 0
2 years ago
Read 2 more answers
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