# Homework Help: Basic Vertical Height Problem

1. Sep 6, 2008

### Eternal Sky

1. The problem statement, all variables and given/known data
A kangaroo jumps to a vertical height of 2.55 m. How long was it in the air before returning to Earth?

2. Relevant equations
v = v(initial) + a*t

x = x(initial) + v(initial)*t + 1/2*a*t^2

v^2 = v(initial)^2 + 2*a*(x - x(initial))

3. The attempt at a solution

I attempted to use the x = x(initial) + v(initial)*t + 1/2*a*t^2 equation, but it didn't work out. I assumed that x and x(initial) were both 0, since the total displacement is 0, and acceleration would be -9.80 m/s^2. However, I don't know what the velocity would be, and I don't see how I can solve the equation without it.

Any help would be appreciated (sorry about the newbie question, but I just started physics)

2. Sep 6, 2008

### LowlyPion

Welcome to PF.

You are almost right, except you are given the height of 2.55 m that it jumps.

If you can figure how long it takes to fall, then you also will know how long it took to get up to the height. That way you can eliminate the need to know Vo.

3. Sep 6, 2008

### HallsofIvy

Notice that x= x(initial)+ v(initial)t+ (1/2)at2 is a quadratic equation and its graph is a parabola. Further, since a= acceleration due to gravity is negative, it is a parabola opening downward. What must v(intitial) be in order that the vertex of the parabola be 2.55 m above x(initial)?

4. Sep 6, 2008

### Eternal Sky

I think I see now.

I set the initial height to 2.55 m and used that to determine the time for half of the trip, which was about 0.72 s. Multiply that by two, and the time for the whole trip is 1.44 s.

Thanks to both of you for your help!