Question

Standing at the edge of a building of height 9.2m, a woman reaches out over the edge and throws a penny vertically upward. It rises, then falls past her, hitting the ground below 2.5s after release. What height did the penny attain? What was its initial velocity?

Answer 1

the equation I am using to find initial velocity
\[v_{f} - v_{i} = at\]
\[v_{i} = v_{f} - at\]
after plugging in for a and t respectively, i get
\[v_{i} = (-9.81)*(2.5)\]
\[v_{i} = 24.525 m/s^{2}\] but my answer is incorrect.
help?

Answer 2

i have realized where I went wrong.
I am using the incorrect general equation to solve for vi

Answer 3

Hello. Here's what I see. You initially assumed the final velocity was 0 m/s so you can find the initial velocity. This would be valid if the time of flight from her hand upward was only 2.5 seconds but the TOTAL time of flight was 2.5 seconds so the time required to leave her hand and go up is less than this. We need to consider our kinematic equations. I suggest\[\Delta y =v _{i} \Delta t + \frac{ 1 }{ 2 } a (\Delta t)^{2}\]
If we arrange this equation for the initial velocity we get
\[v _{i} = \frac{ (\Delta y - \frac{ 1 }{ 2 } a (\Delta t^{2})) }{ \Delta t }\]
Now, the value we will use for (Delta y) will be -9.2m. This is because the ball will fall to the ground a displacement of 9.2 m below the starting point. The acceleration we use will be -9.81 m/s^2 as this the acceleration of gravity on Earth. We also know that (Delta t) is 2.5s. Follow the algebra to find that the initial upward velocity must have been 8.6 m/s (accounting for sig figs.). This is part I. Part II in a sec...

Answer 4

The equation you will use for part II will be \[v _{f ^{2}} = v _{i ^{2}} + 2 a \Delta y\]
If we re-arrange this problem, such that it is solved for the vertical displacement, we obtain\[\Delta y = \frac{ (v _{f ^{2}} - v _{i ^{2}})}{ 2 a }\]
Knowing that the initial velocity upward is +8.6 m/s and the final velocity of the penny at the top will be 0 m/s (as the penny will be still for a fraction of a second while it changes direction to begin falling downward) and also that our value for the acceleration of gravity is -9.81 m/s^2, we can substitute these values in for our variables. In doing so we find that the penny rises a height of 3.8 m above the top of the building before falling down. Thus, the TOTAL height the penny was above the ground will be 3.8 m + 9.2 m (building height), yielding a total height of 13.0 m for the penny above the ground.

Answer 5

Okay, I LOVE your explanation, this is how I try to approach anybody whoasks for help: with as much of a solid explanation as I can give.
Now, I am a bit confused on Part I.
If I solve for vi, I get:
\[v_{i} = \frac{(-9.2m)}{2.5s} - \frac{(9.81m/s^{2})}{2}\]
I get
\[v_{i} = -15.9425\]???? maybe I'm inputting it wrong.

Answer 6

I tried using the formula for conservation of energy but for some reason I'm not getting the same answer.
\[\frac{ 1 }{ 2 }mv _{i}^{2}+mgy _{i} = mgy _{f}\]

Answer 7

yeah I got the answer, I was putting the problem into the calculator incorrectly.