Throwing an Egg Straight Up: Solving for Initial Velocity

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burnst14
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If an egg is thrown straight up in the air, then falls and reaches a point 30m below its starting point 5 seconds after it leaves the thrower's hand:
What is the initial speed of the egg?
How high does it rise above its starting point?
What is the magnitude of its velocity at the highest point? 0 m/s
What are the magnitude and acceleration at the highest point? 9.8 m/s2 downward
Sketch a-t, v-t, and x-t graphs for the motion of the egg. Easy enough

I have tried so many different methods and they all bring me to what seem like dead ends.

I wrote out three equations just for the initial upward movement and tried to substitute:

x=0+v0t+1/2(-9.8)t2
0=v0+(-9.8)t
0=v02+2(-9.8)x

My attempt at substituting:
The first equation becomes: v0=x/t+4.9t
The second becomes: v0=9.8t
The third becomes: v02/19.6=x

Putting the first and second together, we get:
9.8t=x/t+4.9t
4.9t=x/t
4.9t2=x
Then substituting the third:
4.9t2=v02/19.6

Dead end?

I also tried:
v02/19.6=v0t+(-4.9)t2

But I don't even know where to start to solve that.

Any ideas folks?
 
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X = 30+2y
Assuming each side of the aec is equal to distance y. Correct?
 
But that doesn't help because the arc portion velocities aren't in the same direction.
 
Oh! DUH?!? Dang. Got it. Thanks guys or girls.

So since v0 is equal to the velocity on the other downward side of the fall at the zero point:
v0=30/5+4.9(5)
 
Actually I'm just going to start over. That equation is too far in for me to remember the context.
 
burnst14 said:
Actually I'm just going to start over. That equation is too far in for me to remember the context.

Are you familiar with the idea that the area under the "v-t graph" is the displacement?
(and the idea that the acceleration is the slope of the "v-t graph"?)

I find that using the graph to write the equation is usually a pretty straight forward approach.
(I would break it up into two times, [itex]\Delta t_1[/itex] and [itex]\Delta t_2[/itex]; where [itex]\Delta t_1[/itex] is the time when [itex]v=0[/itex], and [itex]\Delta t_1+\Delta t_2=5[/itex] seconds)
(Also, time zero (t=0) is when [itex]v=v_i[/itex])

This may not be the simplest way for everyone, but I just thought I'd tell you about the method that I personally find useful.
 
burnst14 said:
Oh! DUH?!? Dang. Got it. Thanks guys or girls.

So since v0 is equal to the velocity on the other downward side of the fall at the zero point:
v0=30/5+4.9(5)

There is a sign error here and it is what I meant by being mindful of the signs. You have defined the up direction as being positive (you have negative acceleration and the gravitational acceleration is down). Thus, if the egg is going to end up 30 meters below the starting point (which you had put at x = 0), what is x at the end point?
 
<complete solutions are not allowed -- post content deleted>

gneill
PF Mentor
 
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Orodruin said:
I suggest you start from here and use PeroK's hint. Be mindful of signs.

Okay going off of this, I'll go with:
v0=-30/5+4.9(5) = 18.5 m/s

I'm not quite sure why this works though.

The unit analysis works out though... I'm a conceptual thinker, so substituting the equations likely threw me off.

So is that correct?
 
burnst14 said:
Okay going off of this, I'll go with:
v0=-30/5+4.9(5) = 18.5 m/s

I'm not quite sure why this works though.

The unit analysis works out though... I'm a conceptual thinker, so substituting the equations likely threw me off.

So is that correct?

Yes, 18.5m/s is correct.

You say you're a conceptual thinker, well you can certainly sanity check your answer:

If it is thrown up at 18.5m/s, it will take a bit less than 2 seconds to stop and the same to come back to its starting position, traveling 18.5m/s downwards. That's in under 4 seconds.

From there it will fall for more than 1 second, going from 18.5 to about 30m/s, so average of about 25m/s.

And 25m/s for a bit more than a second is about 30m.

So, that's how you could sanity check your answer.