- #1

Argonaut

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- Homework Statement:
- In a Fourth of July celebration, a firework is launched from ground level with an initial velocity of 25.0 m/s at 30.0° 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 ±53.0° with respect to the horizontal, both quantities measured relative to the original firework 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?

- Relevant Equations:
- Basic trigonometry

The official solution says ±25.4°, but I'm having trouble reproducing it. Here is my solution:

1) The components of the velocity of firework F with respect to the ground G in the moment of explosion are the following (Notice, I'm using sin, because the statement says

$$v_{F/G-x}=\sin{30,0^{\circ}} * 25.0 \textrm{m}/\textrm{s}=12.5\textrm{m}/\textrm{s}$$

$$v_{F/G-y}=0$$

2) The components of the velocity of the piece of shrapnel S are the following (for the sake of simplicity, observing only the one flying above the horizontal):

$$v_{S/G-x}=v_{S/F-x}+v_{F/G-x}=(\cos{53,0^{\circ}} * 20.0 \textrm{m}/\textrm{s}) + 12.5\textrm{m}/\textrm{s} = 24.5 \textrm{m}/\textrm{s}$$

$$v_{S/G-y}=v_{S/F-y}+v_{F/G-y}=(\sin{53,0^{\circ}} * 20.0 \textrm{m}/\textrm{s}) + 0=16.0\textrm{m}/\textrm{s}$$

3) Finding the angle:

$$\alpha = \tan^{-1}{(v_{S/G-y}/v_{S/G-x})} = 33.1^{\circ}$$

Am I misunderstanding what 'from vertical' means? Or have I made some other trig error?

1) The components of the velocity of firework F with respect to the ground G in the moment of explosion are the following (Notice, I'm using sin, because the statement says

**30.0° from vertical**.)$$v_{F/G-x}=\sin{30,0^{\circ}} * 25.0 \textrm{m}/\textrm{s}=12.5\textrm{m}/\textrm{s}$$

$$v_{F/G-y}=0$$

2) The components of the velocity of the piece of shrapnel S are the following (for the sake of simplicity, observing only the one flying above the horizontal):

$$v_{S/G-x}=v_{S/F-x}+v_{F/G-x}=(\cos{53,0^{\circ}} * 20.0 \textrm{m}/\textrm{s}) + 12.5\textrm{m}/\textrm{s} = 24.5 \textrm{m}/\textrm{s}$$

$$v_{S/G-y}=v_{S/F-y}+v_{F/G-y}=(\sin{53,0^{\circ}} * 20.0 \textrm{m}/\textrm{s}) + 0=16.0\textrm{m}/\textrm{s}$$

3) Finding the angle:

$$\alpha = \tan^{-1}{(v_{S/G-y}/v_{S/G-x})} = 33.1^{\circ}$$

**Therefore, the two pieces of shrapnel move at ±33.1° with respect to the horizontal and as measured by a spectator standing on the ground.**

Except that the official solution says ±25.4°, which is the value I get if at step 1, I calculate the x component of the velocity of the firework using the cosine and not the sine, so if ##v_{F/G-x}=\cos{30,0^{\circ}} * 25.0 \textrm{m}/\textrm{s}=21.65\textrm{m}/\textrm{s}##. But the statement says 30.0° from vertical.Except that the official solution says ±25.4°

Am I misunderstanding what 'from vertical' means? Or have I made some other trig error?