How can I accurately calculate the duration of a free fall jump?

[ScienceGeek]

A. Here is the stated inquiry:

A man jumps to a vertical height of 2.7 m. How long was he in the air
before returning to Earth?

B. Equations used:

Y = Yo + VoT + (1/2)gT2 (Y because motion is in the vertical direction)

C. Attempted solution:
Because the total distance was twice the man’s highest point (5.4m total) I
used 5.4 rather than 2.7.

Y = Yo + VoT + (1/2)gT2
5.4 = 0 + 0 + ½ (9.8)(t2)
T = 1.05 s

I checked the answer in the back of the book, which was 1.5, but I have been able to come to this solution. Any direction as to what I am doing incorrectly?

 

[Doc Al]

B. Equations used:

Y = Yo + VoT + (1/2)gT2 (Y because motion is in the vertical direction)

Realize that Y is the object's position (or height), not distance traveled.

C. Attempted solution:
Because the total distance was twice the man’s highest point (5.4m total) I
used 5.4 rather than 2.7.

That's your error. The man never reached a height of y = 5.4m!

 

[ScienceGeek]

I attempted calculations using the total displacement as 0, and again as position = 2.7m, but I'm missing something because I just can't get the answer. This seems like it should be much simpler. I'm assuming the initial velocity is zero, but is that an incorrect assumption? Should I be trying to calculate V first?

 

[Doc Al]

I'm assuming the initial velocity is zero, but is that an incorrect assumption?

If the initial speed was zero (at the moment he left the ground) how far would he get?

Express the initial velocity in terms of other variables: acceleration and time. (Use another kinematic formula relating them.)

You can also use symmetry to make life easier: The motion going up is the exact reverse of the motion going down. So you only need to find the time it takes for half the motion and then double it.

 

[ScienceGeek]

Thank you. I was half-way there, just forgot to double the time.