Basic Physics Question please help

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  • #1
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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.
 

Answers and Replies

  • #2
Doc Al
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Start by writing equations that describe the horizontal and vertical displacement as a function of time.
 
  • #3
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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
 
  • #4
Doc Al
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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).
 
  • #5
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how do i find the horizontal velocity component?
 
  • #6
Doc Al
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Use the basic definition of velocity.
 
  • #7
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can i do this:

y = 1/2 * g* t^2
y = 1/2 (9.8) (2)^2
y = 19.6
 
  • #8
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or is that only when an object is thrown directly upwards and falls directly downwards.
 
  • #9
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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
Doc Al
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Well... 40m is the horizontal distance... so that must be the horizontal component of velocity.
 
  • #11
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now how do i find vertical component?
 
  • #12
Doc Al
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Do what I suggested in post #2. What equation describes the vertical motion?
 
  • #13
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y = y0 + v0y*t - 1/2*g*t^2 ?
 
  • #14
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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
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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
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what makes you think its initial velocity is 0
 
  • #17
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its on the ground?
 
  • #18
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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
Doc Al
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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
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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
Doc Al
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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
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so all the - demonstrates is the negative direction?
 
  • #23
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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
Doc Al
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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
Doc Al
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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....
 

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