Throwing objects in the air and calculating height problem

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In summary, the problem involves a man throwing an object straight up at a speed of 10 m/s and the object hitting the ground 5 seconds later. The questions involve determining the highest point reached by the object, the height of the building, and the speed at which the object hits the ground. The initial velocity of 10 m/s will eventually slow down the object due to the force of gravity, until it reaches its highest point where its kinetic energy will be zero and potential energy will be at its highest. The object will then do a free fall back to the ground, with its kinetic energy increasing and potential energy decreasing. To determine the highest point, the condition v_y = 0 can be used, and to determine the
  • #1
BeeGeeks
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Hey folks I'm having some trouble with this question in my homework:

Homework Statement



A man at the top of a building throws an object straight up at a speed of 10 m/s. The object hits the ground 5 seconds latter. For the purpose of this problem the assumed environment is a vaccum.

a) What is the highest point the object reached?
b) What's the height of the building?
c) At what speed does the object hit the ground?



2. The problem

How does the 10 m/s speed relate to the 9.8 m/s^2 gravitational acceleration, does it just slow down the object at -9.8 m/s^2 ?
 
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  • #2
How does the 10 m/s speed relate to the 9.8 m/s^2 gravitational acceleration, does it just slow down the object at -9.8 m/s^2 ?

I may be wrong here but if the initial velocity will as it states the question cause the object to oppose the force of gravity. As the object moves up due to the force that the man threw it up, gravity is acting on it. Eventually gravity will slow the object down to 0 (at its heighest point above the ground where its kinetic energy will be 0 and its potential energy will be its greatest) and start bringing it back to Earth until the object eventually hits the ground and is stopped by the ground.

Just trying to help, hopefully someone can verify my idea. You may wish to take a look at: http://www.physicslearningsite.com/projectile.html
 
  • #3
Hello ,
As the man imparts kinetic energy to the object , The body starts moving up . But as gravity is acting downwards , its KE will start to decrease till it becomes zero , which would be at the highest point . Now at the highest point the particle will do a free fall and reach the ground . During the free fall the KE will increase , and Potential energy due to gravity will decrease .
If you take the regular sign convention , we can take the initial velocity to be positive and gravity as negative , which means the same thing as gravity is in opposite direction to velocity .
 
  • #4
A man at the top of a building throws an object straight up at a speed of 10 m/s. The object hits the ground 5 seconds latter. For the purpose of this problem the assumed environment is a vaccum.

a) What is the highest point the object reached?
b) What's the height of the building?
c) At what speed does the object hit the ground?

To start you off, the object will reach it's highest point when [itex] v_y = 0.[/itex]Using this condition and another kinematic relation, you should be able to find the highest point. (above the point of release)
 
  • #5
For (b)

Total time = 5secs.
Initial velocity up.
For conservation of energy, the magnitue of velocity up is equal to magnitude of velocity down at point s=0

Find the time taken for the object to return to same level as it was thrown, that is s=0.
Then you have time to reach the ground and its inital velocity
 

Related to Throwing objects in the air and calculating height problem

What is the formula for calculating the maximum height of an object when thrown in the air?

The formula for calculating the maximum height of an object when thrown in the air is h = (v^2 * sin^2θ) / (2g), where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

How does the angle of projection affect the maximum height of a thrown object?

The angle of projection affects the maximum height of a thrown object by determining the vertical and horizontal components of its initial velocity. An angle of 90 degrees (straight up) will result in the maximum height, while any other angle will result in a lower maximum height.

What is the acceleration due to gravity on Earth?

The acceleration due to gravity on Earth is 9.8 meters per second squared (m/s^2). This value is constant and is denoted by g in the formula for calculating maximum height.

What factors can affect the accuracy of calculating the maximum height of a thrown object?

The accuracy of calculating the maximum height of a thrown object can be affected by factors such as air resistance, wind, and human error in measuring the initial velocity and angle of projection. Additionally, the formula assumes a flat surface and neglects any external forces acting on the object, which may also affect the accuracy of the calculation.

Can the formula for calculating maximum height be used for all objects thrown in the air?

No, the formula for calculating maximum height is only applicable for objects thrown vertically with a constant initial velocity and under the influence of gravity. Objects with varying initial velocities, angles of projection, or experiencing air resistance will require different equations for calculating their maximum height.

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