How to Calculate the Velocity of a Projected Object at Impact?

  • Thread starter Thread starter ashi1416
  • Start date Start date
  • Tags Tags
    Angles
Click For Summary

Homework Help Overview

The discussion revolves around calculating the velocity of a projected object, specifically a sack of doorknobs thrown into the air with a known initial velocity and angle. Participants are exploring the relationships between the components of velocity and the equations governing projectile motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, including initial velocity and angle, and question how to determine the final velocity upon impact. There are inquiries about the horizontal and vertical components of velocity and how they relate to the overall motion.

Discussion Status

Some participants have provided insights into the components of velocity and the use of gravitational acceleration to find the vertical component at impact. There is an ongoing exploration of the equations needed to arrive at the final answer, with no consensus reached on a specific method yet.

Contextual Notes

Participants are working under the constraints of a homework assignment that requires showing work for the final answer. There is a focus on understanding the physics concepts rather than simply obtaining a numerical result.

ashi1416
Messages
5
Reaction score
0

Homework Statement


a physics student pitches a sack of doorknobs into the air with an initial velocity of 24.5 m/s at 36.9degrees from the horizontal.
i know the:
time=3.0s
horizontal distance traveled=58.8m
time when sack is at max height=1.5s
max height sack reaches=11.03m

but i need to know how to figure:
the velocity of the sack as it hits the ground=?

ok so i have the correct answer its:
24.5 m/s, trajectory is 322.9degrees or 36.9degrees with horizontal


Homework Equations



i need to know how you got this answer
like what equations did you use?

The Attempt at a Solution



i got that far as to figuring out all the questions but how do you get that one?


(my teacher gives us the question and the answer and wants to see our work...thats how i know that last answer)
 
Physics news on Phys.org
ashi1416 said:

Homework Statement


a physics student pitches a sack of doorknobs into the air with an initial velocity of 24.5 m/s at 36.9degrees from the horizontal.
i know the:
time=3.0s
horizontal distance traveled=58.8m
time when sack is at max height=1.5s
max height sack reaches=11.03m

but i need to know how to figure:
the velocity of the sack as it hits the ground=?

ok so i have the correct answer its:
24.5 m/s, trajectory is 322.9degrees or 36.9degrees with horizontal

Homework Equations



i need to know how you got this answer
like what equations did you use?

The Attempt at a Solution



i got that far as to figuring out all the questions but how do you get that one?(my teacher gives us the question and the answer and wants to see our work...thats how i know that last answer)

What are the x and y components of velocity when the sack lands?

[tex]\vec{V} = \vec{V_x} +\vec{V_y}[/tex]
 
x=58.8 and y=0
is that what your asking for??
 
noo
ok but how do you get the Vx and Vy
the Vo=24.5
 
ashi1416 said:
x=58.8 and y=0
is that what your asking for??

What was it horizontal velocity at that time + what was the vertical velocity.

Add them together as vectors. You will get a magnitude and an angle.
 
ashi1416 said:
noo
ok but how do you get the Vx and Vy
the Vo=24.5

You had to have had the horizontal velocity to know how far it traveled.
 
Vy is given by how fast it was going falling from 11 meters.

Or Vy = g*t to fall from that height where t is the same t you found to get to that height.
 
oh ok i have the initial horizontal velocity its 19.59 and the initial vertical velocity is 14.7

do need any final velocitys to solve this?
or am i going off the initial the whole time?
 
ashi1416 said:
oh ok i have the initial horizontal velocity its 19.59 and the initial vertical velocity is 14.7

do need any final velocitys to solve this?
or am i going off the initial the whole time?

The initial Vx will remain the same.

The final Vy will be downward. Since it was at the same height at the end that it started, downward velocity will be the same magnitude as the start, but opposite sign. But note it's only because it's at the same height.
 
  • #10
oh ok.

ok so what like equatitions with the numbers in them would you use to got to the final answer of 24.5m/s?
 
  • #11
ashi1416 said:
oh ok.

ok so what like equatitions with the numbers in them would you use to got to the final answer of 24.5m/s?

The components form two sides of a right triangle. Any way you can think of to calculate the Hippopotamus?
 

Similar threads

Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 46 ·
2
Replies
46
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
4K