Finding the peak height of a projectile with DAVTU?

[Esoremada]

EDIT: I figured it out, the time data was wrong for the trial. redid it with the correct 0.95s and got 1.5m peak height which seems right. :D

Homework Statement

In my physics class we made a miniature trebuchet (the axle is 50cm off ground) and part of the lab report is to calculate the peak height of the projectile given the vertical displacement (height of the arm at release point), horizontal displacement (where the projectile lands) and time from launch to landing.

We measured vertical displacement of .9m, the arm extends about a metre above the ground. The distance and time for the trial are 6.2m and 1.6s

Homework Equations

I used

d = V1t + .5at^2 to find V1

(V2)^2 – (V1)^2 = 2ad to find d when V2 = 0

The Attempt at a Solution

Horizontal velocity (vx):

the projectile went 6.2m in 1.6s

Vx = d/t = 6.2/1.6

Vx = = 3.875m/s
Initial vertical velocity (v1y)

d = V1(t) + at^2

-0.9 = V1(1.6) + (-10)(1.6^2)

-0.9 = V1(1.6) + (-5)(2.56)

-0.9 = V1(1.6) - 12.8

V1(1.6) = 11.9

V1 = 7.4375m/s
Peak height:

Peak dy from launch point:

(V2^)2 – (V1)^2 = 2ad

to find the height at the peak, make v2 = 0 and then isolate d, am I doing this right?

0 - 7.43752 = 2(-10)d

-55.316 = (-20)d

d = 2.7658m

Total peak height from ground:

2.7658 + 0.9 = 3.6658

But a peak height of 3.5m is impossible, where did I go wrong?

 

[jbsteele]

Thanks in advance for your help! The answer is correct, you just had the wrong time data. The correct time data should be 0.95s for the trial, not 1.6s. With that, the peak height of the projectile should be approximately 1.5m.