1 / 9

Projectile Motion

Projectile Motion. y. x. A plane flying at an altitude of 5 km drops a bomb. The speed of the plane is 720 km/h (about 200 m/s). 140 m/s. Once released from the plane, the motion of the bomb is best described as being :

sierra-richard
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

Presentation Transcript

  1. Projectile Motion

  2. y x A plane flying at an altitude of 5 km drops a bomb. The speed of the plane is 720 km/h (about 200 m/s). 140 m/s Once released from the plane, the motion of the bomb is best described as being: • Independent motion in the x and y direction, each with constant velocity • Independent motion in the x and y direction, each with positive uniform acceleration • Independent motion in the x and y direction, with constant velocity (x) and uniform negative acceleration (y). • It isn’t that simple, because the motion is going to be a complicated combination of both x and y directions. 5 km

  3. y x A plane flying at an altitude of 5 km drops a bomb. The speed of the plane is 720 km/h (about 140 m/s). 200 m/s When the bomb is dropped, its trajectory will look like: • (2) (3) (4) 5 km

  4. y 5 km x 200 m/s What is the bomb’s initial velocity when it is dropped from the plane? • [ 200 x + 0 y ] m/s • [ 0 x + 200 y ] m/s • [ 200 x - 9.8 y ] m/s • [200 x – 200 y ] m/s • None of these

  5. y 5 km x 200 m/s To calculate the time require to hit the ground, we can use the kinematic equation for displacement with uniform acceleration, y = vi t + ½ g t2 , using the values: • y= 5, vi = 200 m/s, g = 9.8 m/s2. • y = -5000 m, vi = 0 m/s, g = -9.8 m/s2 • y = 5000 m, vi = 0 m/s, g = -9.8 m/s2 • Can’t use this equation.

  6. y 5 km x 200 m/s Which of the following most likely describes the final velocity vector components of the bomb when it strikes the target on the ground? • vxf = 200 m/s, vyf = 0 m/s. • vyf = something negative, but can’t determine vxf because we don’t know the acceleration. • vxf = 0 m/s, vyf = some negative number. • vxf = 200 m/s, vyf = some negative number.

  7. A cat is sitting on the floor, 0.5 meters from the kitchen counter. Cats always jump so that their vertical velocity component is zero at the exact moment when they reach the target spot. With what velocity and take-off angle must the cat jump if the counter is 1 meter high? 1 m 0.5m

  8. y x A plane flying at an altitude of 5 km drops a bomb. The speed of the plane is 720 km/h (about 200 m/s). 200 m/s • Determine the final velocity vector of the bomb, in component form as well as magnitude+angle. • How long does it take the bomb to fall to the ground? • How far downrange (horizontal distance) from the target must the bomb be dropped? • If the bomber dives such that the bomb has an initial vertical velocity component of -40 m/s, determine the new downrange distance that the bomb must be dropped. 5 km

More Related