This text is adapted from Openstax, University Physics Volume 1, Section 4.3: Projectile Motion. Where θ is the direction of the displacement. Finally, solve for the magnitude and direction of the displacement and velocity using Recombine quantities in the horizontal and vertical directions to find the total displacement and velocity.The problem-solving procedures here are the same as those for one-dimensional kinematics. The air resistance is c v 2 or c v like normal.

If wind is in the direction of motion at 3 m/s and the projectile is moving at 10 m/s then the air resistance term uses 7 m/s.

Note that the only common variable between the motions is time, t. The idea is that you have the air resistance force that depends on velocity and it gets modified to account for wind speed.

Solve for the unknowns in the two separate motions.

Use the kinematic equations for horizontal and vertical motion presented earlier. The equation for the objects height (s) at time (t) seconds after launch is s(t) -4.9t2 + 19.6t + 58.8, where s is in meters.

Treat the motion as two independent one-dimensional motions: one horizontal and one vertical.

The magnitudes of the vertical and horizontal components of velocity are given by v⋅sin( θ) and v⋅cos( θ), respectively, where v is the magnitude of the velocity and θ is its direction relative to the horizontal. The magnitudes of the components of displacement along these axes are x and y.

Initially, resolve the motion into horizontal and vertical components along the x- and y-axes.

For example, a basketball thrown by a player, an arrow shot from a bow, and kids jumping into the pool, all undergo projectile motion.Īny projectile motion problem can be solved by using the following strategy: Label the distances and velocities given in the problem on your picture. Projectile motion is commonly observed in our day-to-day life. Draw out the scenario so you can see how the object travels. Rearranging the terms in x-component equation and substituting it in the equation of y-position, we get the final trajectory equation. Its equation can be obtained by using the position equations. The trajectory describes the projectile's path in a 2-dimensional space. Find parametric equations that describe the motion of the. By substituting the total time in the displacement equation in the x-direction, the equation for the maximum range of the projectile is obtained. Ron throws a ball straight up with an initial speed of 60 feet per second from a height of 5 feet. The total duration of the flight is twice the time of half-flight. The zenith height of the projectile can be obtained by substituting the expression for the half-flight time in the displacement equation for y-direction. By substituting v y in the velocity equation for y-direction, time for half-flight is obtained. The initial velocity components obtained by using trigonometric relations further simplify the motion equations.Īt the zenith, v y is zero. Putting a y as minus g and a x as zero for a projectile, the equations simplify.

The motion of a projectile along the horizontal and vertical directions can be analyzed independently by using the equations of motion.

Projectile equation