How Do You Calculate Initial Velocities for Projectiles?

  • Thread starter Thread starter Oceanspire
  • Start date Start date
  • Tags Tags
    Resultant Velocity
Click For Summary
To calculate the initial velocities for a projectile launched from a 2-meter wall that lands after 2 seconds, the relevant equations include v = at and d = (1/2)at². The acceleration due to gravity is known, allowing for the determination of vertical velocity. The initial upward velocity can be calculated by considering the net change in height and the time of flight. However, the second question regarding the initial resultant velocity lacks clarity, as it likely refers to the velocity just before impact rather than at launch. Proper application of the SUVAT equations is necessary for accurate calculations.
Oceanspire
Messages
4
Reaction score
0
Poster has been reminded to show their Attempt at a Solution
1. You launch a projectile from a wall that is 2 meters in height. It lands on the ground 2 seconds later.

-What is the initial upward velocity of the projectile at the moment of launch?
-What is the initial resultant velocity of the projectile at the moment of launch?2. Equations given:
- v = at
- d = (1/2)at2
- t2 = d/a

I tried using the last equation to find the acceleration, but I can't seem to figure it out at all.
 
Last edited by a moderator:
Physics news on Phys.org
Your third equation disagrees with your second.
You need to show some attempt, as per the template and guidelines.
 
haruspex said:
Your third equation disagrees with your second.
You need to show some attempt, as per the template and guidelines.
This is my homework assignment and these are the equations that were given to me to use but I don't know how. I just need help. I don't know where to start.
 
Oceanspire said:
I tried using the last equation to find the acceleration
Presumably the setting is the surface of the Earth, so you don't need to find the acceleration, it is already known.
It doesn't say, but I assume you are only asked to find the vertical velocity. There's not enough information to find any horizontal component.
The second question makes no sense. Is that the exact wording?
 
haruspex said:
Presumably the setting is the surface of the Earth, so you don't need to find the acceleration, it is already known.
It doesn't say, but I assume you are only asked to find the vertical velocity. There's not enough information to find any horizontal component.
The second question makes no sense. Is that the exact wording?
Yes, except the first question which I'll fix in a second. I forgot to add that I want to find the initial upward velocity
 
Oceanspire said:
Yes, except the first question which I'll fix in a second. I forgot to add that I want to find the initial upward velocity
Ok.
For the first question:
The first thing to do is choose your coordinate system. Which do you want to make the positive direction, down or up? (Up is sort of standard, but it's your choice.)
You know the acceleration (which is?)
You know the time.
What is the net change in height (d)?
The equation you quote, d = (1/2)at2, is not general enough. It assumes starting from rest, which is not the case here. Can you quote a more standard 'SUVAT' equation?

The second question still makes no sense. My guess is they meant to ask for the velocity just before hitting the ground.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
40
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
11
Views
1K
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
26K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
1K
  • · Replies 53 ·
2
Replies
53
Views
5K