- #1

DeltaForce

- 38

- 6

- 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?

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?