AP Calculus/physics help

A rocket is fired vertically upward with an average velocity of 100 ft per second. An observer standing 300 ft from the launch site takes a snapshot when his ground level camera's line-of-sight to the rocket is at an angle of 32 degrees from the horizontal. How many seconds after lift-off did the observer snap the picture and how high was the rocket at this time?

If you could also show me how to do it, that would be a big help to me. Thanks in advance

let us assume the rocket started from point P on the ground and the observer is standing at point O 300 ft from P. let us assume that he took the photo t seconds after the launch. let the rocket be at point Q on space directly above p after t seconds. we see that on joining points O,P &Q, we get a right angled triangle whose one of the angles other than right angle is known to be 32 degrees. the length of OQ is calculated as 100t( distance=speed * time,) now in triangle POQ, OP=300 ft, PQ=100t ft.
we know that tanx=opposite side/adjacent side
from the triangle,
tan (32 degrees) = 100t/300=t/3
ie t= 3 * tan32 =1.874 s
so we found out that the observer took the picture about 1.874 seconds after launch
for finding the height of rocket we just have to multiply the speed of rocket(100 ft/s) with time (1.874 s) ie height = 187.4 feet

This is a trigonometry question.

Use Tan(32) = opp/adj = Height/300

So height = 300 Tan(32) ft

Time = Height/100 seconds