1 / 34

390 likes | 606 Views

Projectile Motion. Objects in projectile motion follow a parabolic path called a trajectory. If released at the same instant, a bullet shot from a gun will hit the ground at the same time as a bullet dropped from the same height.

Download Presentation
## Projectile Motion

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Objects in projectile motion follow a parabolic path called**a trajectory.**If released at the same instant, a bullet shot from a gun**will hit the ground at the same time as a bullet dropped from the same height.**The horizontal and vertical motions of a projectile are**independent, meaning they do not affect each other. Vertical motion Horizontal motion**The 2-D motion of a projectile can be separated into two 1-D**motions: horizontal and vertical.**The horizontal motion of a projectile is always constant, if**we neglect air resistance.**For projectiles shot at 0°, all of the initial velocity is**in the x direction. Thus, Vyi = 0 m/s.**For projectiles shot at 0°, vertical displacement and**velocity will always be negative.**Rules for Projectile Motion**• Treat horizontal and vertical as two separate sides of the problems • TIME is the key, and the only variable that can be used for both horizontal and vertical • Horizontal Motion is always constant • vx is constant • ax = 0 m/s2 • Objects follow a parabolic shape**Horizontal Projectile Motion**• All of the initial velocity is in the x direction, Vyi = 0 m/s • Vertical displacement and velocity will always be negative**For projectiles shot at an angle, initial velocity is both**vertical and horizontal.**Horizontal velocity and initial vertical velocity can be**found using trig functions. Vi Vx = Vi Cos θ Vyi θ Vyi = Vi Sin θ Vx**At the end of the problem, you can recombine horizontal and**vertical velocity to get the total 2-D velocity. Vx Vf Vyf Vx Vf2 = Vx2 + Vyf2 θ Vyf Vf θ = tan-1(Vyf / Vx)**For APM, vy at any height is the same while going up and**coming down except for direction.**Velocities of projectile motionNote: Vy = 0 at the highest**point.**Angled Projectile Motion**• Initial velocity is both vertical and horizontal • Use trig functions to find vyi and vx • Vx = Vi Cos θ • Vyi = Vi Sin θ • Remember clues • vy at the top is 0 m/s • vy at any height is the same while going up and coming down except for direction**Example Problem**Happy Gilmore hits his shot at 55.0 m/s with an angle of 50.0° to the ground. How far did the ball travel before it lands? • vi = 55.0 m/s • θ = 50.0° • ay = -9.81 m/s2 • ∆x = ?**vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ?θ = 50.0°**Find vx and vyi • vx = vi Cos θ = 55.0 m/s Cos (50.0°) = 35.4 m/s • vyi = vi Sin θ = 55.0 m/s Sin (50.0°) = 42.1 m/s**vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ?θ = 50.0°**vyi = 42.1 m/s vx = 35.4 m/s What do we need to find ∆x? Time! Find time from the vertical side ∆y = vyi∆t + ½ ay∆t2 0 m = (42.1 m/s) ∆t + ½ (-9.81 m/s2) ∆t2 - (42.1 m/s) ∆t = ½ (-9.81 m/s2) ∆t2 (42.1 m/s) = ½ (9.81 m/s2) ∆t ∆t = 8.58 s**vi = 55.0 m/s ay = -9.81 m/s2 ∆x = ?θ = 50.0°**vyi = 42.1 m/s vx = 35.4 m/s∆t = 8.58 s Now we can find ∆x ∆x = vx∆t = (35.4 m/s)(8.58 s) = 304 m**Where Should We Aim the Cannon?**At or Above the monkey?**That’s how the space shuttle and satellites orbit the**earth.

More Related