Time vs Height - EASY but easily forgotten :/

[Synix]

1.Homework Statement
One day, while working on a downtown construction site, Eugenia accidentally dropped her cell phone into the large excavation on the site. Fortunately, the cell phone landed on a pile of soft Earth and was not damaged. She asked Tony, the construction manager, who was working nearby to throw the cell phone back up to her.

The height of an object thrown vertically up is modeled by

s(t) = s0 + v0t - 1/2gt^2

where s0 is the initial height above the ground, v0 is the initial velocity, g is the acceleration due to gravity, t is the time, and s(t) is the height above the ground at time t.

If Tony is 10m below ground level, and the acceleration due to gravity is 9.8m/s^2, with what initial velocity must he throw the cell phone for it to reach Eugenia? Explain any assumptions you make and show your calculations, graphs, or both to justify your answer.

Homework Equations

h(t) = t^2 + t + c

s(t) = s0 + v0t - 1/2gt^2

The Attempt at a Solution

Fail.
http://i72.photobucket.com/albums/i169/Alkapwn/math2-1.jpg

 

[Gib Z]

Hello Synix, welcome to PF!

Are you familiar with the equation $$v_f^2 = v_i^2 + 2a\Delta s$$ ?

Thats the equation we need for this one. After Tony throws up the cellphone, when it reaches Eugenia it should be at the top of its flight. At this point, was is its velocity? I'm sure you can figure it out from here.

 

[Architeuthis Dux]

I can understand how easily it is to forget certain equations or concepts, especially when they are not used frequently. However, it is important to always double check and refresh our memory before attempting to solve a problem. In this case, the equation s(t) = s0 + v0t - 1/2gt^2 is the correct one to use, as it models the height of an object thrown vertically up.

In order to solve this problem, we need to make some assumptions. First, we will assume that Tony throws the cell phone with an initial velocity (v0) directly upwards, as this is the most efficient way to get the cell phone back to Eugenia. Second, we will assume that the initial height (s0) of the cell phone is 10m, since Tony is 10m below ground level. Finally, we will use the known value for the acceleration due to gravity, which is 9.8m/s^2.

Now, we can plug in our values into the equation s(t) = s0 + v0t - 1/2gt^2 and solve for v0.

s(t) = 10m + v0t - 1/2(9.8m/s^2)t^2

Since we want the cell phone to reach Eugenia, we can set the final height (s(t)) equal to 0.

0 = 10m + v0t - 1/2(9.8m/s^2)t^2

Solving for v0, we get:

v0 = 4.9t^2/t - 10m/t

Since t is the time it takes for the cell phone to reach Eugenia, we can estimate this value to be around 1 second. Therefore, v0 = 4.9m/s - 10m/s = -5.1m/s.

In conclusion, Tony must throw the cell phone with an initial velocity of -5.1m/s in order for it to reach Eugenia. It is important to always double check our assumptions and equations to ensure accurate and precise solutions.