Maximum Velocity in projectiles

In summary, the problem involves a man standing on a 16.6m high building and throwing a rock with a velocity of 25.0m/s at an angle of 35.0 degrees above the horizontal. By ignoring air resistance, the magnitude of the velocity of the rock just before it strikes the ground can be calculated using the equations for projectile motion. The y-component of the velocity is found to be 23m/s and, by considering conservation of energy, the x-component can also be determined. The impact angle may not necessarily be the same as the launch angle due to the different elevations of the impact and launch points.
  • #1
seanpk92
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Homework Statement


A man stands on the roof of a building of height 16.6 and throws a rock with a velocity of magnitude 25.0 at an angle of 35.0 above the horizontal. You can ignore air resistance.

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Homework Equations



y - y0 = v0y2 + 1/2 ayt2

vy2 = v02 + 2ay(y - y0)



The Attempt at a Solution



I calculated that the projectile would go 10.5 meters above the building which would make the total height of 27.1m. I also calculated how much time it would be in the air - 3.74s - and then found the range - 76.6m (I didnt do all that for fun, it was to answer other parts of this question). I figured using the second equation I listed would give me the velocity of the y component.
I get vy = 23m/s and then divided by sin(35) to get the velocity

?
 
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  • #2
The impact angle will not necessarily be the same as the launch angle, because impact is happening at a different elevation than the launch point.

You've determined the y-component of the velocity, and you should be able to determine the x-component. Add them like you would any vector components.

[By the way, you could also get to the answer by employing conservation of energy]
 
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Related to Maximum Velocity in projectiles

1. What is maximum velocity in projectiles?

Maximum velocity in projectiles is the highest speed that a projectile can reach during its flight. It is typically measured in meters per second (m/s) or feet per second (ft/s).

2. How is maximum velocity calculated in projectiles?

Maximum velocity in projectiles is calculated using the equation V = √(2gh), where V is the velocity, g is the acceleration due to gravity (9.8 m/s² or 32 ft/s²), and h is the height of the projectile's launch point above the ground.

3. What factors affect the maximum velocity in projectiles?

The maximum velocity in projectiles is affected by several factors, including the initial launch speed, the angle of launch, air resistance, and the force of gravity. The shape and weight of the projectile can also have an impact on its maximum velocity.

4. Is there a limit to the maximum velocity in projectiles?

Yes, there is a limit to the maximum velocity in projectiles. This is because as the projectile travels further, it experiences air resistance and drag, which gradually slows it down. Eventually, the projectile will reach its terminal velocity, the maximum speed it can attain given the air resistance and gravity acting upon it.

5. How does maximum velocity in projectiles affect their trajectory?

The maximum velocity in projectiles has a direct impact on their trajectory. A higher maximum velocity will result in a longer and flatter trajectory, while a lower maximum velocity will result in a shorter and steeper trajectory. Additionally, the angle of launch and air resistance also play a role in determining the trajectory of a projectile.

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