1 / 16

Horizontal Projectile Motion

Horizontal Projectile Motion. Brian Chen & Jomir Uddin. What is Horizontal Projectile Motion?. -When an object is launched horizontally it follows a parabolic path that goes downwards.

chiquita-lane
Télécharger la présentation

Horizontal 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

Presentation Transcript

  1. Horizontal Projectile Motion Brian Chen & Jomir Uddin

  2. What is Horizontal Projectile Motion? -When an object is launched horizontally it follows a parabolic path that goes downwards. -Horizontal projectile motion refers to the trajectory of an object thrown straight forward, with no initial vertical component. -There is only an initial horizontal component.

  3. What are the components of a horizontally moving object? -The velocity vector of a horizontally moving object points towards its path of motion -Although there is no initial vertical velocity, due to gravity, g, there is vertical velocity while the object is in flight. -Gravitational acceleration causes gradual increase in vertical velocity. As a result, two objects dropped from the same height will hit the ground at the same time

  4. Horizontal Velocity Stays Constant! Air resistance is neglected AND there is no horizontal acceleration in projectile motion!

  5. Range -Range is the amount of distance that the object travelled after being tossed. -Range is dependent on the initial horizontal velocity (which stays constant) and the time of flight. -Use the following to solve for various variables dx = vixt -When only height from which the object was thrown is given, one must calculate the time of flight.

  6. Range Pt. 2 The time of flight is the time that it takes the object to hit the ground. To calculate the range, when no time is given, use the formula: -g is gravitational acceleration, which is 9.81m/s^2 -d is the height from which the object is thrown

  7. Question Number One A pitcher throws a baseball a distance of 18.5 meters to a batter. If it takes 0.65 seconds for the ball to reach the batter, at what velocity was the ball thrown?

  8. Answer Number One WRONG dx=vt v=dx/t v=(18.5m)/0.65s v=28.5m/s

  9. Question Number Two A plane carrying crates of bananas is moving east with a horizontal velocity of 300m/s. Unfortunately, one of these crates was not properly strapped in place, and fell out of the airplane. What is the horizontal velocity of the crate right before it strikes the ground?

  10. Answer Number Two 300m/s! -Remember, the horizontal velocity remains constant! Air resistance is neglected!

  11. Question Number Three A lead pipe is thrown horizontally from one of the windows of a 20 story building. If it lands 15 meters from the foot of the building in 5 seconds, from what story was the lead pipe thrown? Each story is 10 meters in height.

  12. Answer Number Three dy=(1/2)gt^2 dy=(1/2)(9.81m/s^2)(5s)^2 dy=(4.905m/s^2)(25s^2) dy=122.625m 120m/10m=12 stories The lead pipe was thrown from the 12th story of the building.

  13. Question Number Four A boy is on a cliff that is 50m tall. He kicks a ball off the cliff with a horizontal velocity of 20m/s. How far does the ball land?

  14. Answer Number Four Remember! When not given the time, use the formula: d = 1/2xgxt^2 50 = 1/2x9.81m/s^2xt^2 3.19s = tdx=vt dx=20m/s x 3.19s dx = 64m

  15. Question Number Five A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

  16. Answer Number Five Use d = 1/2 gt^2 to solve for time; the time of flight is 2.12 seconds. Now use dx = vix t - 1/2 at^2 Note that a is 0 m/s^2By substituting 35.0 m for dx and 2.12 s for t, the vix can be found to be 16.5 m/s.

More Related