To find the resultant speed of the boat, we must represent the velocity of the boat and the current as vectors. We will then add the two vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

The boat is traveling east at , represented as a vector 

The current is traveling north at , represented as a vector 

Adding the two vectors together, gives us the resultant velocity of the boat:

The resultant speed of the boat is the magnitude of the velocity.

If a vector is in the form , its magnitude,  ,its found by using the equation 

Given the velocity vector:

The resultant speed of the boat is:

Definition of momentum:

Combine equations:

Plug in values:

A picture of the pumpkins motion is shown above. We wish to find the range that the pumpkin travels. We align our axes so that the x-axis points in the direction of the horizontal motion of the pumpkin, with x=0 being the x position of the cannon head and the y-axis points vertically off the ground. Therefore the pumpkin's intial position is  and . Our pumpkin's initial velocity is at  to the horizontal. We want the x and y components of the velocity, and that is found by resolving the  vector into x and y-components given by 

 and  

Since we want the final x position of the pumpkin, we write out our kinematics equation for position in the x-coordinate: 

Since there is no air resistance, , and substituting  in for  and  for  gives us 

However, we still don't know how long the pumpkin is in the air. But we do know that it's final y-position . Therefore we will use our kinematics equation for position in the y-coordinate to find the time. We have:

Since this is a projectile motion problem, we know that , and we already know that , , and . Plugging these in to the y-position equation gives us:

This is a quadratic equation that we can solve for time by using the quadratic formula. Doing so will give us two values: 

 and 

Since time is always positive, we use the value . Thus the pumpkin flies in the air for this long. Now we can plug this into our x-position equation to find . Doing so gives:

. Therefore the range of the projectile is .

The above diagram illustrates the situation. Originally, the meteor (shown on the left) is traveling with some initial momentum. And on the right the two chunks are shown splitting off. Each must have half the original mass as given in the problem, so the masses are each , but the velocities are unknown. Since no forces are acting on the system, momentum is conserved, and we can apply the conservation of momentum for the system. We split the conservation of momentum into two directions, one along the original trajectory which we will call the x direction, and one normal to the original trajectory which we will call the y direction. Then our conservation of momentum equations are  and . For our initial momentums, we see that the meteor has momentum in the x-direction but no momentum in the y-direction, therefore: 

 and 

For the final momentum, we see that both meteors have momentum in the positive x-direction, and opposite momenta in the y-direction. Therefore we have: 

Therefore, our equations of conservation become:

To make things simpler, we divide through by the mass  to get:

Now we have two equations in two unknowns,  and . Since we want the velocity of the meteor that split at , we want to find . Therefore, we solve the second equation for  to obtain . Now we substitute this value into the first equation to obtain:

Factoring out the  and using the trigonometric identity , we get:

Now we divide each side by the expression in the brackets to obtain , and we get:

Therefore the speed of the chunk flying out at  is .

In this question, we're told that a projectile (the clown) is launched at a certain velocity and a certain angle. We're asked to find the maximum height of the projectile.

We need to recognize that this is a question concerning motion in two dimensions, both x and y. This means that the projectile will be launched such that it travels in a parabolic path. However, for the purposes of this question, we can neglect motion in the x direction and instead direct our attention to the projectile's vertical motion.

Since the projectile travels in a projectile path, we know that at the instant it reaches its maximum height, its velocity will fall to zero. We can take advantage of this fact, combined with one of the kinematic equations (since we know acceleration due to gravity is constant), in order to find our answer.

And since we know that gravity is oriented in the downward direction, we can give it a negative value.

Now, we need to find an expression that gives us the clown's initial velocity in the vertical direction. Remember, the question stem gives us the clown's initial velocity, but this is not the same as its velocity in the y direction. The value that is given in the question stem is a combination of both the x direction velocity as well as the y direction velocity. Thus, we'll need to use the information about the clown's initial velocity provided in the question stem in order to calculate the initial vertical velocity.

Now that we have an expression for the initial vertical velocity, we can go ahead and plug this value into the prior expression that we derived.

Finally, if we plug in the known values, we can calculate our answer.

Since we are asked for the horizontal distance, we need to calculate the horizontal component of the arrow's initial velocity. Also, since we are neglecting air resistance, we know that this velocity is constant. Therefore, we simply need to multiply it by the time interval given in the problem statement to calculate the distance.

Now we can multiply this by the time interval given, which is 0.5 seconds:

The arrow will reach its maximum height when the vertical component of its velocity is 0. Therefore, we will first calculate the vertical component of the arrow's initial velocity:

Now we can calculate how long it takes for this velocity to equal 0

First, we need to calculate the vertical component of the initial velocity of the arrow:

Then we can use the following kinematic equation to calculate the change in height of the arrow:

Rearranging for change in height:

The arrow reaches its maximum height when the vertical component of its velocity is 0.  Plugging in values for each variable:

Now adding this to the initial height we get:

We can use the range equation to calculate how far the arrow travels until it gets back its initial height. However, we then need to calculate how far the arrow travels from there until it hits the ground. So, beginning with the range equation:

Plugging in our values:

Now we need to calculate the horizontal distance that the arrow travels after this point. First, let's calculate how long it takes to hit the ground. At this point, the arrow has the same speed that it began with, but its velocity is now directed at  below the horizontal. Therefore, its vertical component:

Then using the following kinematic equation:

Plugging in our values:

Rearranging:

Then using the quadratic formula, we get:

 or 

Obviously, we can't have a negative time, so the first option is the correct one. We can then multiply this by the horizontal component of the arrow's velocity, which is constant since we are neglecting air resistance, to get the rest of the horizontal distance traveled:

Then adding our two distances together to get our total distance traveled, we get:

The key component of the problem statement is that the arrow is shot from a height of 1m and hits a target that is also at 1m. Therefore, since we are neglecting air resistance, we can conclude that the arrow hits the target at the same speed that it begins with, except that the velocity is now oriented  below the horizontal. Thus, the arrow hits the target with the same kinetic energy that it begins with, and the target absorbs all of this energy. Therefore, we can say:

Note that with the information given in the problem statement, the angle that the arrow was shot from and the distance the target was from where the arrow was shot are both insignificant.