Batman (mass = 90 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +20 m/s. What is the velocity of the boat after Batman lands in it?

The total momentum for motion along the water does not change, assuming the water applies negligible retarding friction along that direction as the boat stabilizes. Therefore

M * 20 = (M + m)V
V = M*20/(M+m)
= (530/620)*20 m/s = ?

Well, it sounds like Batman is really going all out in his crime-fighting efforts! But let's not forget that he's also a man, not a superhero with actual superpowers. Anyway, let's calculate the velocity of the boat after Batman jumps in.

To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the jump must be equal to the total momentum after the jump.

The initial momentum of the boat is given as its mass (530 kg) multiplied by its initial velocity (+20 m/s), which is 10,600 kg*m/s.

The momentum of Batman can be calculated based on his mass (90 kg) and his initial velocity (which is 0 since he's jumping straight down). So his momentum is 0 kg*m/s.

Since momentum is conserved, the total momentum after the jump should be 10,600 kg*m/s.

Now, the total mass after the jump is the sum of the boat's mass (530 kg) and Batman's mass (90 kg), which is 620 kg.

Finally, we can calculate the velocity of the boat after Batman lands by dividing the total momentum (10,600 kg*m/s) by the total mass (620 kg).

10,600 kg*m/s ÷ 620 kg ≈ 17.1 m/s

So, the velocity of the boat after Batman lands in it is approximately 17.1 m/s. It seems like Batman's landing made quite a splash!

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before Batman jumps is equal to the total momentum after Batman lands in the boat.

Before Batman jumps, the boat has a momentum of:

momentum_boat_before = mass_boat * velocity_boat
= 530 kg * 20 m/s
= 10600 kg⋅m/s

Batman's momentum before he jumps is:

momentum_batman_before = mass_batman * velocity_batman
= 90 kg * 0 m/s (Batman is initially at rest)
= 0 kg⋅m/s

The total momentum before Batman jumps is simply the sum of the boat's and Batman's momenta:

total_momentum_before = momentum_boat_before + momentum_batman_before
= 10600 kg⋅m/s + 0 kg⋅m/s
= 10600 kg⋅m/s

After Batman lands in the boat, the boat and Batman move together with a final velocity, which we'll call v_f.

The total momentum after Batman lands is:

total_momentum_after = (mass_boat + mass_batman) * v_f

According to the conservation of momentum principle, the total momentum before is equal to the total momentum after:

total_momentum_before = total_momentum_after

10600 kg⋅m/s = (530 kg + 90 kg) * v_f
10600 kg⋅m/s = 620 kg * v_f

Hence, the equation for the final velocity, v_f, is:

v_f = 10600 kg⋅m/s / 620 kg
v_f ≈ 17.1 m/s

Therefore, the velocity of the boat after Batman lands in it is approximately +17.1 m/s.

To determine the velocity of the boat after Batman lands in it, you can use the principle of conservation of linear momentum.

The conservation of linear momentum states that the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system.

In this case, the "system" includes both Batman and the boat. Before Batman lands, the boat has an initial velocity of +20 m/s. After Batman lands, the boat will have a new velocity, which we need to calculate.

To calculate the velocity of the boat after Batman lands, we can use the following equation:

(mass of Batman * velocity of Batman before landing) + (mass of boat * velocity of boat before landing) = (mass of Batman * velocity of Batman after landing) + (mass of boat * velocity of boat after landing)

Plugging in the values we know:

(90 kg * 0 m/s) + (530 kg * 20 m/s) = (90 kg * velocity of Batman after landing) + (530 kg * velocity of boat after landing)

0 + (10,600 kg m/s) = (90 kg * velocity of Batman after landing) + (530 kg * velocity of boat after landing)

10,600 kg m/s = (90 kg * velocity of Batman after landing) + (530 kg * velocity of boat after landing)

Since we want to find the velocity of the boat after Batman lands, we can rearrange the equation as follows:

(530 kg * velocity of boat after landing) = 10,600 kg m/s - (90 kg * velocity of Batman after landing)

Now, we have an equation that relates the velocity of the boat after landing to the velocity of Batman after landing. To solve for the velocity of the boat, we need to know the velocity of Batman after landing.

If no external forces are acting on the system, meaning no external forces are slowing down the boat or Batman, we can assume that the momentum is conserved, and the equation becomes:

(530 kg * velocity of boat after landing) = 10,600 kg m/s - (90 kg * velocity of Batman before landing)

Now, we need to know the velocity of Batman before landing. That information is not provided in the question, so we cannot determine the velocity of the boat after Batman lands without that information.