Solving Kinematics Problem: V_o & Max Height

[DeltaForce]

Homework Statement

John climbs a tree to get a better view of the speaker at an outdoor graduation ceremony. Unfortunately, he leaves his binoculars behind. Marsha throws them up to John, but her strength is greater than her accuracy. The binoculars pass John's outstretched hand after 0.75s and again 1.28s later. How high is John?

Relevant Equations

v_o + at = v_f
v_o*t - 0.5gt^2 = y

For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.

Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation.

9.947(0.75) - 0.5(9.8)(0.75^2) = 4.70m

I checked the answer and solution, but it was totally different from my answer. What did I do wrong?

 

[berkeman]

How high is John?

Obviously we need more information to answer this question. What was he smoking before climbing the tree?

For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.

Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation.

9.947(0.75) - 0.5(9.8)(0.75^2) = 4.70m

I checked the answer and solution, but it was totally different from my answer. What did I do wrong?

Check the wording of the question. The 1.28s is *after* the 0.75s first cross-over time...

 

[DeltaForce]

Hahahaha, I was sitting there wringing my brain out figuring out what I did wrong. Now I see. Thank you!

 

[berkeman]

Happy to help. I try to avoid climbing trees after, well, you know...