How Do You Calculate the Initial Velocity Components of a Projectile?

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Homework Help Overview

The discussion revolves around calculating the initial velocity components of a projectile given its horizontal and vertical displacements after a certain time. The problem involves understanding projectile motion, specifically the relationships between displacement, velocity, and acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations of motion for horizontal and vertical displacements, questioning the assumptions about initial velocities and the effects of gravity. There are attempts to derive the initial velocity components based on given displacements and time.

Discussion Status

Participants are actively engaging with the problem, offering equations and discussing the implications of their assumptions. Some have calculated the horizontal component of velocity, while others are working on the vertical component. There is a recognition of the need to clarify the role of gravity and the initial velocity in the equations.

Contextual Notes

There is an ongoing debate about the initial conditions of the projectile, particularly whether the initial vertical velocity can be assumed to be zero. Participants are also navigating the implications of using negative signs in their calculations related to gravity.

kara
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2 seconds after being projected from ground level, a projectile is displaced
40 m horizontally and 53 m vertically above its launch point. What are the horizontal and vertical components of the initial velocity of the projectile? At the instant the projective ahieves maximum height above ground level, how far is it displaced horizontally from the launch point?

Any suggestions for that questions, I've already drawn a diagram.
 
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Start by writing equations that describe the horizontal and vertical displacement as a function of time.
 
so would saying the following make any sense:

the horizontal velocity is constant so the projectile is moving 40m/s right
the vertical velocity changes 9.80 m/s downward because of gravity
 
kara said:
the horizontal velocity is constant so the projectile is moving 40m/s right
The horizontal velocity component is constant, but it's not 40 m/s.
the vertical velocity changes 9.80 m/s downward because of gravity
The vertical velocity component accelerates (due to gravity) at 9.8 m/s^2 downward; acceleration is the rate of change of velocity (m/s per s).
 
how do i find the horizontal velocity component?
 
Use the basic definition of velocity.
 
can i do this:

y = 1/2 * g* t^2
y = 1/2 (9.8) (2)^2
y = 19.6
 
or is that only when an object is thrown directly upwards and falls directly downwards.
 
alright, i know that v = d/t so i'd get v = 40m/2sec = 20 m/sec but do i use the horizontal distance or vertical distance ? and what does velocity give me? is that the horizontal component?
 
  • #10
Well... 40m is the horizontal distance... so that must be the horizontal component of velocity.
 
  • #11
now how do i find vertical component?
 
  • #12
Do what I suggested in post #2. What equation describes the vertical motion?
 
  • #13
y = y0 + v0y*t - 1/2*g*t^2 ?
 
  • #14
so:

- initial position (y0) would be 0 m
- initial velocity (v0) would be 0 m/s
- time would be 2.0 sec
- force of gravity 9.8 m/s^2

so my answer would be -19.6 m/s??
 
  • #15
wait.. wouldn't gravity be -9.8 m/s^2 so that the answer would be 19.6 ?

and i don't know what units it'd be
 
  • #16
what makes you think its initial velocity is 0
 
  • #17
its on the ground?
 
  • #18
The velocity you calculated is if you dropped it off a cliff (in which case its initial velocity would be zero). However, in this case it is projected upwards meaning it was given some initial velocity
 
  • #19
kara said:
so:

- initial position (y0) would be 0 m
OK.
- initial velocity (v0) would be 0 m/s
v0 is what you are trying to find! Don't assume it's anything--certainly not zero. You know it moved up so v0 can't be zero!

- time would be 2.0 sec
OK
- force of gravity 9.8 m/s^2
The acceleration due to gravity is -9.8 m/s^2.

You left out one piece of data: the vertical displacement at time t.

Solve for v0.
 
  • #20
Doc Al said:
OK.

v0 is what you are trying to find! Don't assume it's anything--certainly not zero. You know it moved up so v0 can't be zero!


OK

The acceleration due to gravity is -9.8 m/s^2.

You left out one piece of data: the vertical displacement at time t.

Solve for v0.


take into consideration his equation has a built in negative so if he put g in as negative it would make the object fly upwards
 
  • #21
BishopUser said:
take into consideration his equation has a built in negative so if he put g in as negative it would make the object fly upwards
Right... don't put g as negative! g is always a positive number; the acceleration due to gravity is -g. (Which is why your equation has the minus sign already.)
 
  • #22
so all the - demonstrates is the negative direction?
 
  • #23
you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?
 
  • #24
kara said:
so all the - demonstrates is the negative direction?
Sure, but that makes a lot of difference. The generic kinematic formula (for constant acceleration) is:
y = y0 + v0t + 1/2at^2

When the acceleration is due to gravity (like this problem) we put a = -g:
y = y0 + v0t -1/2gt^2
 
  • #25
kara said:
you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?
Exactly...
 
  • #26
you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?
 
  • #27
alright so solving that equation would give me v0. Now what do i do with that? would i also need to solve for v0 in equation x=x0 + v0t?
 
  • #28
:O I just figured something out... Don't know if I am on the right track but let me know:

If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?
 
  • #29
Solving that equation gives you the vertical component of the initial velocity; you already solved for the horizontal component.
 
  • #30
Alright, so i got 36.3 m/s for v0
 

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