What Angle Does the Rocket Appear to Move at After the Second Stage Ignites?

[student34]

Homework Statement

A toy rocket takes off from the ground with an initial velocity of 25.0m/s 60.0° from the horizontal. Once it reaches its maximum height, the second stage of the rocket kicks in at 20.0m/s 53.0° from the horizontal. From an observer on the ground, what angle would the rocket appear to be moving from the horizontal immediately after the second stage started?

This question only takes into account velocities and nothing else more complicated.

Homework Equations

I just used cosθ = x/r, sinθ = y/r and tanθ = y/x

The Attempt at a Solution

I began with Vx(total) = 25.0m/s*cos(60°) + 20.0m/s*cos(53°) = 24.536m/s.
Then Vy(total) = 20.0m/s*sin(53°) = 15.97m/s; I only have one Vy component because it said that the rocket reached its maximum height from the first stage of the rocket.

Then, I had θ = tan^(-1)[(15.97m/s)/(24.536m/s)] = 33.1°, but my book' answer is 25.4°
I really think it's wrong, but can anyone else see how it could be 25.4°?

 

[haruspex]

Your answer is correct. One way to get the book answer is to confuse horizontal and vertical velocity components in the first stage (equivalently, to take the initial trajectory as 60 degrees to the vertical).

 

[student34]

Your answer is correct. One way to get the book answer is to confuse horizontal and vertical velocity components in the first stage (equivalently, to take the initial trajectory as 60 degrees to the vertical).

Ah, you're right. That explains their answer, thanks!