- #1
burnst14
- 53
- 2
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?
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?