Finding launch angle and velocity

In summary, the conversation discusses the problem of launching a rock over an enemy castle wall, which is 12 meters high and surrounded by a moat. The distance of 17 meters from the wall must be taken into account. The conversation delves into finding the necessary angle and velocity to successfully clear the wall at the top of the projectile's trajectory. Through a series of calculations and solutions, it is determined that the required initial vertical velocity is approximately 15.34 m/s and the required horizontal velocity is approximately 10.897 m/s. These two components can then be used to find the complete velocity and solve for the angle needed to successfully launch the rock over the wall.
  • #1
Forrest
9
0

Homework Statement


You are an invading army who wants to launch a rock over the enemy castle’s wall. The wall is 12 meters high and there is a moat surrounding the wall which forces you to launch from a distance of 17 meters away. What angle and velocity should you shoot at in order to just clear the wall at the top of your projectile’s trajectory?

Homework Equations


No idea.

The Attempt at a Solution


I have read through the relevant chapter in my book twice now trying to find a solution or process to find the angle and velocity for a specific height, but all I can find is referring to maximum distance. I would really appreciate any help even just pointing me in the right direction!

Thanks so much in advance!
 
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  • #2
Forrest said:

Homework Statement


You are an invading army who wants to launch a rock over the enemy castle’s wall. The wall is 12 meters high and there is a moat surrounding the wall which forces you to launch from a distance of 17 meters away. What angle and velocity should you shoot at in order to just clear the wall at the top of your projectile’s trajectory?

Homework Equations


No idea.

The Attempt at a Solution


I have read through the relevant chapter in my book twice now trying to find a solution or process to find the angle and velocity for a specific height, but all I can find is referring to maximum distance. I would really appreciate any help even just pointing me in the right direction!

Thanks so much in advance!
Hello Forrest. Welcome to PF.

In the future please show an attempt at a solution or at an understanding.

What must the vertical component of the (initial) velocity be for a projectile to reach a height of 12 meters?
 
Last edited:
  • #3
SammyS said:
Hello Forrest. Welcome to PF.

In the future please show an attempt at a solution or at an understanding.

What must the vertical component of the (initial) velocity be for a projectile to reach a height of 17 meters?

My apologies. Wouldn't the initial vertical velocity have to be about 3.33? I could do the problem given either a velocity or an angle but I am extremely lost on how to do it without both. Thanks!
 
  • #4
Forrest said:
My apologies. Wouldn't the initial vertical velocity have to be about 3.33? I could do the problem given either a velocity or an angle but I am extremely lost on how to do it without both. Thanks!
How did you get 3.33 ?
 
  • #5
SammyS said:
How did you get 3.33 ?
Whoops i mistyped. I meant 15.33, but that was the lowest value I found that could give a maximum vertical height of 12 meters.
 
  • #6
Forrest said:
Whoops i mistyped. I meant 15.33, but that was the lowest value I found that could give a maximum vertical height of 12 meters.
Which does not answer the question ... How do you find that velocity?
 
  • #7
SammyS said:
Which does not answer the question ... How do you find that velocity?
I graphed a generic polynomial like Ax^2 +Bx+C, and since I know the A value is -4.9 I plugged that in, and I know that the object is starting from the ground so I made C equal to 0 and I just played with the value of B until I got 12 meters in height. I'm sorry if I'm being difficult, my professor has assigned an entire homework on this type of problem and did not explain at all how to solve anything other than a very basic example.
 
  • #8
Forrest said:
I graphed the a generic polynomial like ax^2 +bx+c, and since I know the a value is -4.9 I plugged that in, and I know that the object is starting from the ground so I made C equal to 0 and I just played with the value of B until I got 12 meters in height. I'm sorry if I'm being difficult, my professor has assigned an entire homework on this type of problem and did not explain at all how to solve anything other than a very basic example.
Yes, 15.34 m/s is approximately correct.

With that as the vertical component of velocity, how much time does it take the projectile to reach that height?
 
  • #9
SammyS said:
Yes, 15.34 m/s is approximately correct.

With that as the vertical component of velocity, how much time does it take the projectile to reach that height?
It would take 1.56 seconds i believe.
 
  • #10
Forrest said:
It would take 1.56 seconds i believe.
Yes.

So that should tell you what horizontal component of velocity is needed to reach the wall (a horizontal distance of 17 meters away) in 1.156 1.56 seconds ?
 
  • #11
SammyS said:
Yes.

So that should tell you what horizontal component of velocity is needed to reach the wall (a horizontal distance of 17 meters away) in 1.156 seconds.
The horizontal velocity would need to be 14.71 m/s then?
 
  • #12
Forrest said:
The horizontal velocity would need to be 14.71 m/s then?
Sorry, I had a typo .

17 meters in 1.56 seconds
 
  • #13
SammyS said:
Sorry, I had a typo .

17 meters in 1.56 seconds
That would make it 10.897 m/s.
 
  • #14
Forrest said:
That would make it 10.897 m/s.
Now you have the two components of velocity needed. Right ?
 
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  • #15
SammyS said:
Now you have the two components of velocity needed. Right ?
And now I construct a right triangle to find the complete velocity and solve for theta to give me the angle?
 
  • #16
Forrest said:
And now I construct a right triangle to find the complete velocity and solve for theta to give me the angle?
Yes.
 
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  • #17
SammyS said:
Yes.
Thank you so much for your help! I was having trouble with those first couple steps, I really appreciate you breaking it down for me. You're great! :)
 

Related to Finding launch angle and velocity

What is launch angle and velocity?

Launch angle and velocity refer to the initial angle and speed at which an object is propelled into the air. These measurements are important in understanding the trajectory of the object and predicting its flight path.

Why is it important to find the launch angle and velocity?

Finding the launch angle and velocity of an object is crucial in many scientific fields, such as physics and engineering. It allows us to accurately predict the motion of the object and understand factors like gravity, air resistance, and projectile motion.

How do scientists determine the launch angle and velocity?

Scientists use various methods to determine the launch angle and velocity of an object. These include mathematical calculations, experiments, and simulations. The specific method used depends on the type of object and the available resources.

What factors can affect the launch angle and velocity?

The launch angle and velocity of an object can be affected by several factors, including the initial force or energy used to propel the object, the mass and shape of the object, and external forces such as air resistance and wind.

What are some real-world applications of finding launch angle and velocity?

Finding the launch angle and velocity has many practical applications, such as in sports like baseball and golf to improve player performance. It is also used in space exploration to launch rockets and satellites, and in military applications for projectile weapons and missile systems.

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