- #1

- 18

- 0

## Homework Statement

A rocket is launched at an angle of 53 above the horizontal with an initial speed of 75 m/s. It moves for 25s along it's initial line of motion with an overall acceleration of 25 m/2

^{2}. At this time its engines fail and the rocket proceeds to move as a freebody.

a) What is the rocket's maximum altitude?

## Homework Equations

displacement = V

_{1}time + (1/2)a(time)

^{2}

V

^{2}= V

^{1}+ 2adisplacement

## The Attempt at a Solution

I've googled this question after attempting it, and have seen several different methods from my own.

For part a, I'm interpreting it as the maximum vertical displacement. I set up a vector diagram, where the angle between the average velocity vector and horizontal velocity vector is 53. I use sin ratio to solve for the opposite, which is sinTHETA x hyp. This gives me 60m/s. I do the same process to find the vertical acceleration, which is 20.0 m/s

^{2}. I plug them into the displacement equation, solve to get 7750m.

Now for the free body motion. Velocity will equal 0 when it gets to it's top. I use the second equation to rearrange for displacement. V2 is 0, V1 is -60

^{2}, acceleration is -9.81. Essentially, displacement is a negative value. Meaning the max. altitude is still 7750m.

Makes sense to me, but I applied the same approach to a different question (identical) with diff. values. The answer was way off. (I did this to see if my approach was right).

What am I doing wrong?