# Finding the peak height of a projectile with DAVTU?

1. Jan 10, 2012

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

I used

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

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

3. 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:

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?

Last edited: Jan 10, 2012