A 41.2-kg boy, riding a 2.57-kg skateboard at a velocity of +5.82 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.12 m/s, 9.72° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

To find the skateboard's velocity relative to the sidewalk, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jump should be equal to the total momentum after the jump.

Let's denote the skateboard's velocity relative to the sidewalk as V_skateboard and the boy's velocity relative to the sidewalk as V_boy.

Before the jump:
The boy's momentum is given by M_boy * V_boy, where M_boy is the mass of the boy and V_boy is the boy's velocity relative to the sidewalk.
The skateboard's momentum is given by M_skateboard * V_skateboard, where M_skateboard is the mass of the skateboard and V_skateboard is the skateboard's velocity relative to the sidewalk.

After the jump:
The boy's momentum is given by M_boy * V_jump, where V_jump is the boy's velocity relative to the sidewalk just after leaving contact with the skateboard.
The skateboard's momentum is given by the product of its mass and velocity relative to the sidewalk at this instant.

Using the conservation of momentum principle, we can write the equation:
M_boy * V_boy + M_skateboard * V_skateboard = M_boy * V_jump + M_skateboard * V_new

We are given the following values:
M_boy = 41.2 kg (mass of the boy)
V_boy = +5.82 m/s (boy's velocity relative to the sidewalk)
M_skateboard = 2.57 kg (mass of the skateboard)
V_jump = 6.12 m/s (boy's velocity relative to the sidewalk just after leaving the skateboard)
V_new = ?

Let's substitute these values into the equation and solve for V_skateboard:

41.2 kg * 5.82 m/s + 2.57 kg * V_skateboard = 41.2 kg * 6.12 m/s + 2.57 kg * V_new

Now, isolate V_skateboard:

(41.2 kg * 5.82 m/s) - (41.2 kg * 6.12 m/s) = -(2.57 kg * V_new) + (2.57 kg * V_skateboard)

V_skateboard = ((41.2 kg * 5.82 m/s) - (41.2 kg * 6.12 m/s) + (2.57 kg * V_new)) / 2.57 kg

Now, we need the value of V_new. We know that the boy's velocity just after leaving contact with the skateboard is 6.12 m/s at an angle of 9.72° above the horizontal. We need to resolve this velocity into horizontal and vertical components.

Horizontal component of V_new:
V_horizontal = V_jump * cos(theta)
V_horizontal = 6.12 m/s * cos(9.72°)

Vertical component of V_new:
V_vertical = V_jump * sin(theta)
V_vertical = 6.12 m/s * sin(9.72°)

Now we have the horizontal and vertical components of V_new. We can use these to find the overall magnitude of V_new using Pythagorean theorem:

|V_new| = sqrt((V_horizontal^2) + (V_vertical^2))

Now substitute this value of V_new into the equation for V_skateboard to get the final answer.