Check the picture below.
how do we know? well, notice h(t), starts off at 12, up up up reaches 47.84 then down down down, which is pretty much the trajectory of a flying object, by the time it gets to 44, is still going down.
now, let's look at g(t), starts off at 10, and goes up up up, never down, by the time it gets to 41, is still going up,
so at second 2, h(t) is 44 and going down, g(t) is 41 and going up, at 2.2 h(t) is 40.16, and g(t) is 44.1, between that lapse, h(t) became 44, 43, 42, 41, in the same lapse g(t) became 41, 42, 43, 44, so somewhere in those values h(t) = g(t).
what does the solution mean? It's the seconds or the instant lapse when the first cannon ball was at the same height as the second cannonball.
Answer:
can you posted a pic of the question or please specify what is the question. for us to help you properly. thanks
This a pretty typical right triangle trig problem; the first step is to figure out what we have and what we want in relation to an acute angle in the problem.
Here we have a right triangle, G=90°, and we're given angle F=23°. So we have to name everything in relation to F.
31 = FG is <em>adjacent </em>to F.
x = GE is <em>opposite </em>to F.
OK, we have an opposite and adjacent; that tells us we need to use the tangent of F. Let's write it:
tan 23° = tan F = opp/adj = x/31
Solving,
x = 31 tan 23°
I hate the calculator part. I used to love that part.
x = 31 tan 23° ≈ 13.16 feet
Answer: 4) x ≈ 13.2 ft
What do you mean
Could you please insert an equation or a problem
