How to Calculate Time in Air for a Trebuchet Projectile?

AI Thread Summary
To calculate the time a trebuchet projectile spends in the air, it's essential to consider both vertical and horizontal components of the initial velocity. The projectile's maximum height was correctly calculated at different angles, but the time of flight requires using the vertical component of the initial velocity, which is found by multiplying the initial speed by the sine of the launch angle. The formula V = u + at should apply to the vertical motion, using the vertical initial velocity instead of the total initial speed. Acknowledging this distinction allows for accurate calculations of time spent in the air, which is crucial for determining the projectile's range. Understanding the separation of vertical and horizontal motion is key to solving projectile motion problems effectively.
Marghk
Messages
20
Reaction score
0

Homework Statement



I'm looking at trebuchets. Basically, a trebuchet shoots a projectile with an initial velocity of 30ms-1 at an angle of 55, 65 and 75 degrees. Friction/Wind resistance do not account.

I need to find the maximum height reached, the time the projectile spends in the air and the range. I am having troubles with the time spent in air, therefore affecting my last calculation.

I got a maximum height of 30.811m at 55 degrees.
37.71m at 65 degrees.
42.842m at 75 degrees.


Homework Equations



As mentioned, I'm having trouble finding the total time spent in air. I know I have he following variables. I know that the time to reach max height x2 is the time of flight.

Y = Verticle, X = Horizontal

Uy= 30ms-1
Vy = 0ms-1 (Obviously no movement at the exact highest point)
a= -9.8ms-2
Sy = 30.811m


I tried using the rule V = u+at

V=u+at
T=(v-u)/a

This gives me (0-30)/-9.8
Giving me 3.06 seconds and a total time of 6.122 seconds.


The Attempt at a Solution



If I wanted to use that equation to find out the time taken if launched at 65 and 75 degrees, I would get the same answer due to the fact that verticle displacement isn't accounted.

I've been going over this again and again and I can't figure out anything.
 
Last edited:
Physics news on Phys.org
Well actually there is movement at the highest point.
It has vertical and horizontal speed. The vertical speed is zero at the highest point, but the horizontal speed is constant.
You said that "Y = Verticle, X = Horizontal", but you never seem to notice the horizontal.
When the angle is 55 for example, it has a vertical and a horizontal component.
Vy = sin 55 * v
Vx = cos 55 * v
So Uy is not 30ms-1.
I think you can solve it now.

EDIT: I just noticed that the values for maximum height are correct. So that's what you did (used vertical and horizontal components).
But then why are you using 30 m/s for (0-30)/-9.8 ? Shouldn't it still be sin 55 * 30?
You are only doing calculations vertically.
The acceleration only acts vertically, the final speed is zero only vertically, so then why use the resultant speed in the formula. You have to use the initial vertical speed as well.
 
Last edited:
husky88 said:
You are only doing calculations vertically.

Thanks, I just realized I wasn't calculating both horizontal and vertical attributes. Can't believe I missed that!

Thanks again :D
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top