Calculate total time taken for an object thrown at an angle from a height

In summary, the conversation discusses the calculation of the total time of flight of a cannon ball that is fired from a cliff with a velocity of 30m/s at an angle of 35°. Two methods are presented, with the first method involving calculating the time taken for the cannon ball to reach its maximum height and then adding that to the time taken for it to hit the ground. The second method involves placing the cannon on the ground to see the full parabola, but it is determined that this method is not feasible.
  • #1
LovePhys
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0

Homework Statement


A cannon ball is fired with a velocity of 30m/s at an angle of 35° from the top of a cliff. The height of the cliff is 50m. Calculate total time of flight of the cannon ball before it hits the ground.

[itex] v_o=30m/s [/itex]
[itex] \alpha=35° [/itex]
[itex] g=10m/s^2 [/itex]


Homework Equations



The equation of the parabola expressing the orbit of the cannon ball:
[itex] y= \frac {-gx^2}{2v_o^2cos^2\alpha} + tan(\alpha) x [/itex]


The Attempt at a Solution



Obviously, a part of the parabola is missing as the cannon ball is struck from the cliff. I chose the position of the cannon as the origin and put a coordinate system there.
I have 2 ways of solving this problem:

1/ Calculate the time taken for the cannon ball to reach the maximum height:

[itex] t=\frac {v_osin(\alpha)}{g} ≈ 1.7 (s) [/itex]
The maximum height is: [tex] h=\frac{v_o^2 \sin^2(\alpha)}{2g}≈15(m) [/tex]

Then, pretending that the cannon ball is struck vertically from the height of 15+50=65(m), the time taken for it to hit the ground is:
[tex] \sqrt{\frac{2.65}{g}}≈3.6(s) [/tex]
So that the total time is about: 1.7+3.6=5.3(s)

2/ I want to place the cannon at a position that allows me to see the full parabola. I worked out the quadratic equation of the parabola, and I also had the new horizontal range of the cannon ball if the cannon is at a new position. However, here is where it get stuck. I do not know the time taken for the cannon ball to get from the new position to the original position, also the new angle at which it was thrown, and the new velocity.

Can anyone please help me with this?

I really appreciate your help!

Thanks in advance!
LovePhys
 
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  • #2
Hi LovePhys! :smile:

Your first method is fine, but slightly quicker might be to use the standard constant acceleration equations in the x and y directions (s = ut and s = ut + 1/2 at2) for the whole trajectory :wink:
LovePhys said:
2/ I want to place the cannon at a position that allows me to see the full parabola. I worked out the quadratic equation of the parabola, and I also had the new horizontal range of the cannon ball if the cannon is at a new position. However, here is where it get stuck. I do not know the time taken for the cannon ball to get from the new position to the original position, also the new angle at which it was thrown, and the new velocity.

I don't really understand what you're trying to do here. :confused:

I don't think any such method is feasible.
 
  • #3
Hello tiny-tim,

Thank you very much for your help and sorry for the late response!

I am sorry that I did not explain my second method really clearly.

The cannon is 50m above the ground, so a part of the trajectory is missing (in contrast with the fact that we can see the whole parabola if the cannon had been put on the ground). My idea is that I want to move the cannon on the ground so that I can calculate the total time of flight of the cannon ball. However, when I decide to do this, I realize that I am missing the information about angle of striking, and the cannon ball's initial velocity, as well as the time taken to reach the original position on the cliff, which is 50m above the ground.

I am struggling with this and hope that you can give me a helping hand.

Thank you!
LovePhys
 
  • #4
LovePhys said:
However, when I decide to do this, I realize that I am missing the information about angle of striking, and the cannon ball's initial velocity, as well as the time taken to reach the original position on the cliff, which is 50m above the ground.

the only way you can find these things is to solve the original question first!
 
  • #5
icist

I would like to commend you on your attempt to solve this problem using two different methods. Both methods are valid and will lead to the same answer. However, I would suggest using the first method as it is simpler and more straightforward.

To answer your question about the second method, you can use the conservation of energy principle to determine the new velocity and angle of the cannon ball at the new position. From there, you can use the same equation for the parabola and solve for the total time of flight.

Overall, your approach to this problem is correct and shows good understanding of the concepts involved. Keep up the good work!
 

Related to Calculate total time taken for an object thrown at an angle from a height

1. How do you calculate the total time taken for an object thrown at an angle from a height?

To calculate the total time taken for an object thrown at an angle from a height, you can use the formula: t = (2 * v * sinθ) / g, where t is the total time, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

2. What is the initial velocity in the formula for calculating the total time?

The initial velocity refers to the speed at which the object is thrown. It can be measured in meters per second (m/s) or kilometers per hour (km/h).

3. How is the angle of launch determined in the formula?

The angle of launch, θ, is the angle at which the object is thrown from the horizontal. It can be measured in degrees or radians.

4. What is the value of g in the formula for calculating the total time?

The value of g varies depending on the location on Earth, but it is generally accepted to be 9.8 m/s² or 32.2 ft/s². It represents the gravitational acceleration towards the center of the Earth.

5. Can the formula for calculating total time be used for any object thrown at an angle?

Yes, the formula can be used for any object thrown at an angle, as long as the initial velocity and angle of launch are known. However, it assumes that there is no air resistance and that the object is thrown on a flat surface.

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