*62. A 40.0-kg boy, riding a 2.50-kg skateboard at a velocity of +5.30 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 6.00 m/s, 9.50° 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.

Question

*62. A 40.0-kg boy, riding a 2.50-kg skateboard at a velocity of +5.30 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 6.00 m/s, 9.50° 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.

Answer 1

Your question contains its own answer.

"..the boy's velocity relative to the sidewalk is 6.00 m/s, 9.50° above the horizontal."

The components of the velocity are:
6.00 cos9.5 = 5.92 m/s horizontal and
6.00 sin9.5 = 0.99 m/s vertical

Answer 2

(6-5.5)cos(7.3)

Answer 3

Two diamonds begin a free fall from rest from the same height, a time ?t (delta t) apart. How long after the first diamond begins to fall will the two diamonds be a distance d apart? Give your answer in terms of the variables given and g. Must be given in variables.. this is a symbolic problem

Answer 4

To find the skateboard's velocity relative to the sidewalk at this instant, we can use the law of conservation of momentum.

Let's assume the positive direction is towards the right.

The initial momentum before the boy jumps can be calculated by multiplying the mass of the boy (40.0 kg) by his initial velocity (5.30 m/s):

Initial momentum = (40.0 kg) × (5.30 m/s) = 212 kg·m/s (to the right)

After the boy jumps, he acquires a new velocity (6.00 m/s) at an angle of 9.50° above the horizontal relative to the sidewalk. To calculate the horizontal component of this velocity, we use trigonometry:

Horizontal component = 6.00 m/s × cos(9.50°) ≈ 5.791 m/s

Since the boy and skateboard were initially at rest and remain in contact, there is no external force acting on them. Therefore, the law of conservation of momentum dictates that the total momentum before the boy jumps is equal to the total momentum after the boy jumps.

Total momentum before = Total momentum after

(40.0 kg + 2.50 kg) × V_skateboard_before = (40.0 kg + 2.50 kg) × 5.791 m/s

Simplifying the equation:

42.5 kg × V_skateboard_before = 42.5 kg × 5.791 m/s

Now we can solve for V_skateboard_before:

V_skateboard_before = 5.791 m/s (to the right)

Therefore, the skateboard's velocity relative to the sidewalk at this instant is 5.791 m/s to the right.