- #1
chudd88
- 23
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Homework Statement
In a Fourth of July celebration, a firecracker is launched from ground level with an initial velocity of 25.0 m/s at 30' from the vertical. At its maximum height it explodes in a starburst into many fragments, two of which travel forward initially at 20.0 m/s at \pm53.0' with respect to the horizontal, both quantities measured relative to the original firecracker just before it exploded. With what angles with respect to the horizontal do the two fragments initially move right after the explosion, as measured by a spectator standing on the ground?
Homework Equations
Hardly any, really. Basic trig.
The Attempt at a Solution
First, the italics in the problem description are included in the book, and are really the source of my trouble. I feel like this is a simple problem, and that the solution in my book is wrong, but it's more likely that I'm misreading the problem.
My approach to this is that when the firecracker explodes, it will have 0 vertical velocity, and the same horizontal velocity it had when it was launched. From there, it's a simple matter of adding to that vector the additional velocity as a result of the explosion, which adds more horizontal and vertical velocity.
The problem I have is the wording of the initial angle of the projectile. The book emphasizes that it's 30' from the vertical. To me, that means that I start at 90', then drop 30'. So, it should be the same as saying 60' above the horizontal. I've found that when I do this problem with an initial angle of 60' above the horizonal, I get the wrong answer. But when I do it at 30' above the horizontal, I get the right answer. I'm trying to understand if the book just confused this point, or if I'm doing something wrong.
So, let's assume that the initial angle is 60' above the horizontal. I will use that for my calculations.
At the time it bursts:
Vx = cos(60) * 25.0 m/s = 12.5 m/s
Vy = 0 m/s
The velocity of the projectiles with respect to the firecracker:
Vx = cos(53) * 20.0 m/s = 12.03 m/s
Vy = sin(\pm53) * 20.0 m/s = \pm15.97 m/s
Adding the vectors, I find that the velocity with respect to the ground is:
Vx = 12.5 m/s + 12.03 m/s = 24.53 m/s
Vy = 0 m/s \pm 15.97 m/s = \pm15.97 m/s
Taking the arctan of (\pm15.97/24.53) gives a final answer of \pm 33.1' with respect to the horizontal.
The problem is the book's answer is \pm25.4'.
Struggling to understand this, I just tried doing the problem again, this time assuming the book's description of the initial angle of the firecracker was wrong. When I do the same steps for an initial angle of 30' above the horizontal, I get 25.4' as the answer.
So, is my book wrong, or is my interpretation of the initial angle incorrect?
Thanks.