- #1

fiftybirds

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## Homework Statement

I had to calculate initial and final velocities for a rocket launch. Data are as follows:

horiz displacement = 17.5m

time = 3.5 s

launch angle = 40 deg

angle to apex from 20m = 11 deg

net vertical displacement = 0

## Homework Equations

I used the following equations:

d = vit + (a[t^2])/2

vfy^2 = viy^2 + 2ad

## The Attempt at a Solution

calculations for initial velocity

horizontal

dx = vixt + (aav[t^2])/2

17.5 m = 3.5s(vix) + (-9.81 m/s^2[3.5s^2])/2

77.6 m/3.5 s = vix

22.2 m/s = vix

vertical

0m = viy(3.5s) + (-9.81 m/s^2[3.5s^2])/2

60.1 m/3.5s = viy

viy = 17.2 m/s

then final velocity

vfy^2 = viy^2 + 2ady

= (17.2 m/s)^2 + 2 (-9.81m/s^2)(0)

= -17.2 m/s ?

I used d = vit -(a[t^2])/2 to calculate this again and got the same answer.

It doesn't seem right that the final velocity would be the same magnitude as the initial velocity... is this answer correct? It also doesn't make sense that we had to measure the launch angle and angle to the apex if they aren't required to solve the problem

Also, since horizontal velocity is constant, wouldn't the final horizontal velocity be the same as the initial horizontal velocity?

Thanks.