Consider an airplane that normally has an airspeed of 120 km/h in a 75 km/h crosswind blowing from west to east. Calculate its ground speed when its nose is pointed north in the crosswind.

Ok, we'll assume that the crosswind completely pushes the plane 75 km/h off course.

To find the answer, we need to do vector addition.

We need to calculate the resultant vector found by adding the normal speed with the speed of the crosswind.

Notice that the three vectors form a right triangle, with the resultant vector acting as the hypotenuse . We'll make use of the Pythagorean Theorem to solve for the magnitude of the resultant speed...

We can assert that we know what the firefighter's acceleration would be if friction were not present. He would fall at 9.8 m/s^2 (as would any object in free-fall). But, due to the friction on the pole, he accelerates only at 4.9 m/s^2--half of the maximum value. The force of friction essentially de-accelerates the firefighter by 4.9 m/s^2, found by the calculation...

9.8m/s^2 - 4.9m/s^2 = 4.9m/s^2

The force of friction is found by taking this accleration and multiplying it by the mass of the firefighter...

A ball is thrown with enough speed straight up so that it is in the air several seconds. What is its velocity .50 s before it reaches its highest point?

The acceleration of the ball will be 9.8m/s^2 (as it would be for any object). Because as the ball goes up and up until it reaches its highest point, it will be slowing down, then we'll write the acceleration as a negative number...

a = -9.8 m/s^2

Recall that acceleration is defined as the change in velocity over a given time interval. We know from experience that a ball thrown straight up into the air, when it reaches its highest point, that it will momentarily stop before it changes direction and begins to fall. Since the ball comes to rest at the highest point we'll call this velocity, velocity 2 (v2), and assign it a value of 0 m/s.

We want to know the velocity 0.50 s before the highest point. We'll call this velocity, velocity 1 (v1), since it occurs before velocity 2 (v2).

So, taking acceleration to be given by the equation...

a = (v2 - v1) / t

..where t is the time interval...

we can rearrange to solve for v1...

- v1 = (a * t) - v2

...multiplying both sides by -1 yields...

v1 = - (a * t) + v2

Now substitute the values a = -9.8m/s^2 t = 0.50 s v2 = 0 m/s

A canoe is paddled at 2.0 km/h directly across a river that flows at 3.0 km/h, as shown in the figure. What is the resultant speed of the canoe relative to the shore?