Set 5 Problem number 11

The horizontal component of the velocity of a projectile is 6.02 meters per second. The vertical component is 5.86 meters per second.

Aided by a sketch, find the magnitude and angle of the velocity of the object with horizontal.

Solution

We could sketch a triangle to depict 1 second's worth of motion, which will be 6.02 meters horizontally and 5.86 meters vertically. The horizontal leg of the triangle will represent a displacement vector of 6.02 meters; the vertical leg will represent a displacement vector of 5.86 meters, which will be placed with its initial point at the terminal point of the first.

The legs of the triangle represent 6.02 and 5.86 meters, and the hypotenuse represents the displacement in 1 second, which indicates the velocity of motion.

The velocity will be the magnitude of this vector, in meters, divided by 1 second, giving velocity in m/s.

The angle of the hypotenuse with horizontal is tan^-1( 5.86 / 6.02) = 44.22 degrees. This is the direction of motion.

We could equivalently just sketch a similar triangle whose legs represent the 6.02 m/s and 5.86 m/s velocities. The hypotenuse will represent the velocity of the projectile.

In this case the length of the hypotenuse will represent the speed of the projectile, in m/s.
The angle will represent the angle of the velocity with horizontal.

Generalized Solution

We regard the horizontal and vertical velocities vx and vy as the velocity components vx and vy of a velocity vector.

The magnitude of the velocity vector will be

v = `sqrt(vx^2 + vy^2)

and its direction as measured from the positive x direction will be

tan^-1(vy / vx), or tan^-1(vy / vx) + 180 deg if vx is negative.

Explanation in terms of Figure(s), Extension

The figure below depicts the velocity components vx and vy and the vector velocity resulting from the combination of these horizontal and vertical velocities.

The resultant velocity is depicted by the vector comprising the hypotenuse of the triangle whose legs are vx and vy.