Projectile motion apple throw problem

[gneill]

Your equation looks okay. You've written it several times. Here it is again:
-4m = (2.86 m/s) t + \frac{1}{2} (-9.8 m/s^2) t^2
Collecting onto one side:
\frac{1}{2} (9.8 m/s^2) t^2 - (2.86 m/s) t - 4m = 0

Can you identify and isolate the coefficients for plugging into the quadratic formula? a=?, b=?, c=?

 

[mackenzieb]

if i am correct..
a=-9.8 (not sure if this is supposed to be -, but i believe so)
b=2.86
c=-4

 

[gneill]

Take a closer look at the equation in its 'one sided' form. What exactly is multiplying t²? What sign does the t term have?

 

[mackenzieb]

there is no negative sign, therefore 9.8 should remain positive.
making the variables:
a=9.8
b=2.86
c=-4

 

[gneill]

there is no negative sign, therefore 9.8 should remain positive.
making the variables:
a=9.8
b=2.86
c=-4

What happened to the 1/2 also multiplying the t² term? And the t term is negative!

 

[mackenzieb]

does that make the a term -4.9? (9.8 divided by 1/2)

 

[gneill]

does that make the a term -4.9? (9.8 divided by 1/2)

Nope. The t² term is positive. So it's 4.9 (actually, you should probably carry an extra decimal place or so through these calculations in order to avoid rounding errors creeping into the results).

The t term is negative.

 

[mackenzieb]

okay so to make sure I am onthe right track then...
a=4.906
b=2.86
c=-4

-4m=(2.86m/s)t+1/2(-9.8m/s^2)t^2
1/2(9.8m/s^2)t^2-(2.86m/s)t-4m=0
-2.86 +/- sqrt 2.86^2-2(4.906)(4)/2(-4/906)

is that the correct setup?

 

[gneill]

Not quite. The t term is negative! b = -2.86.
And the quadratic formula is
x = \frac{-b ± \sqrt{b^2 -4ac}}{2a}

See: http://en.wikipedia.org/wiki/Quadratic_equation

 

[mackenzieb]

so when i use that equation...i get

t=-2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)

 

[gneill]

so when i use that equation...i get

t=-2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)

Well, the algebra as written is a bit ambiguous, but that's the general idea. What result do you get?

 

[mackenzieb]

when i get to the final step...
-2.86+/-sqrt-2.86^2-4(4.906)(-4)/2(4.906)
i get -2.86+/- 2.677007231

i am unsure what to do with the -2.86+/-

 

[gneill]

Your casual mathematical grammar is making it difficult to determine where your problem lies. I can't tell what your intention is when you have open-ended square roots and dangling divisions. Please make more use of parentheses to delimit the mathematical operations. If you choose not to use Latex to write your mathematics, then at perhaps you can at least treat the square root as a function: sqrt( stuff ), with brackets to delimit the argument.

The quadratic formula provides TWO roots to a given quadratic equation. The ± indicates how to alter the given formula to produce the two roots. While both roots may satisfy the mathematical requirements of the equation, only one may satisfy the physical constraints (things like time being a positive quantity, or distance a positive value).

 

[mackenzieb]

here is my complete work so far:


Given:
d=4.0m [down]
a=9.8m/s^2 [down]
vi=0m/s (whenever something falls from a height, the v1 is always 0)

Required:
t
vf

Therefore, I use the equation
d=vi(t)+1/2a(t)^2
4m [down] = 0m/s(t)+1/2 (9.8m/s^2[down])(t^2)
4m[down]=1/2(9.8m/s^2[down])(t^2)

Because the initial velocity (vi) is 0 the equation can be simplified...

t=sqrt 4.0m[down]/1/2(9.8m/s^2 [down])
t=0.9035 s

My question is, where am I going wrong ?

 

[gneill]

There's nothing wrong other than this is not the same problem that was given in the first post of this thread! There it was stated that the object was thrown with an initial upward angle of 35°.

 

[mackenzieb]

Yes that is right, the object was thrown with an initial upward angle of 35 degrees. I am trying to use the examples in my notebook I have been provided with, and that it what is making this so difficult.

I didnt think the angle had to be applied until the second question, but i must be incorrect.

I am unsure of how to incorportate the angle into my answer (ie i don't know where it needs to be placed in the answer and how to apply it )

 

[gneill]

Yes that is right, the object was thrown with an initial upward angle of 35 degrees. I am trying to use the examples in my notebook I have been provided with, and that it what is making this so difficult.

I didnt think the angle had to be applied until the second question, but i must be incorrect.

I am unsure of how to incorportate the angle into my answer (ie i don't know where it needs to be placed in the answer and how to apply it )

The elevated launch angle means that there will be an initial velocity in the y-direction.

 

[mackenzieb]

I don't understand what you mean by that
so I do need to incorporate the angle into this question
...meaning the initial velocity is still 0 but is 35 degrees in the y-direction?
I don't know how I am supposed to place this into the equation

 

[gneill]

I don't understand what you mean by that
so I do need to incorporate the angle into this question
...meaning the initial velocity is still 0 but is 35 degrees in the y-direction?
I don't know how I am supposed to place this into the equation

The launch speed is 5 m/s as stated in the problem. The launch is at an angle to the horizontal, so that this speed will have a horizontal and a vertical component. What are they?

The initial velocity components plug into the equations of motions that pertain to the vertical and horizontal motions.

 

[MazharB]

Gneill when i use the quadratic form, and plug in the given variables i do not get 1.27 or 1.23 or anywhere near that
∆t = -2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)
t=-3.8 or -1.91

 

[gneill]

Gneill when i use the quadratic form, and plug in the given variables i do not get 1.27 or 1.23 or anywhere near that
∆t = [[/color]-2.86+/- sqrt( -[/color]2.86^2-4(4.906)(-4)[/color]][/color]/([/color]2(4.906) )[/color]
t=-3.8 or -1.91

The 'b²' term inside the square root should not be negative. The 'c' term is not negative in this case, it's the initial height which is positive.

Initial height is +4m. Final height is 0.00m (ground).

The kinematic equation in the vertical direction is

0 = h + v_y t - \frac{1}{2}g t^2

 

[MazharB]

thanx a million i know see were i went wrong yup its equal to 1.23 sec
for b i used dx=(Vx)(t)
Vx=5.0 m/s x Cos35=4.09m/s
dx=(4.09m/s)(1.23s)
dx= 5.03m
for c i did V^2=(5.0m/s)^2+2g(4m)
V= 10.16 m/s
i just can't figure out how to get the angle,i know I am supposed to use the tan function to find the angle i plug in 10.16*tan(35)=7.14 hmm this does not make sense? were am i going wrong?

 

[gneill]

You have found the speed of the apple when it reaches the ground (that's its speed along its direction of motion). To find the angle that this speed makes with respect to some reference direction (which direction do you think would be appropriate?) you should draw a picture showing this speed along the hypotenuse of a triangle along with another component that you know as one of the sides. Solve for whatever angle is appropriate.