vertical speed observed = -37.9 sin 50.8
= -29.4 m/s
how far did it fall to reach -29.4 m/s?
v = -g t
-29.4 = -9.81 t
t = 2.99 s
d = 4.9 t^2
= 4.9 (2.99)^2 = 43.9
so it reached 43.9 + 22.5 = 66.42 m
vertical speed at ground?
fell from 66.42 m
66.42 = 4.9 t^2
t = 3.68 s
v = -g t = -9.81 (3.68) = -36.1 m/s
additional time = 3.68-2.99
hits ground with initial velocity up reversed and same initial horizontal speed
u = 36.1 /tan 50.8 = 29.5
so speed at ground = sqrt u^2+v^2)
= sqrt (29.5^2+ 36.1^2)
= 46.6 m/s
A small rocket is fired in a test range. It rises high into the air and soon runs out of fuel. On the way down it passes near an observer (sitting in a 22.5-m-high tower) who sees the rocket traveling at a speed of 37.9 m/s and moving in a vertical plane at an angle of 50.8° below horizontal.
What is the maximum altitude of the rocket?
With what speed does the rocket hit the ground?
Once the rocket passes the observer, how long does it take to hit the ground?
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