1 - 1.) Projectile: At t = 0, an object is projected with a...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1.) Projectile: At t = 0, an object is projected with a speed v0 = 45 m/s at an angle q0 = 21.8° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive. For parts a-g, calculate the requested quantities at t = 10 s into the flight. (Use 9.81 m/sec2 for g.) a) The vertical acceleration of the object: ay = m/s2 * -9.81 OK b) Its horizontal acceleration: ax = m/s2 * 0 OK c) Its vertical velocity: vy = m/s * -81.38 OK d) Its horizontal velocity: vx = m/s e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees: angle = ° f) Its vertical displacement, from where it started: y = m g) Its horizontal displacement, from where it started: x = m h) At what time does the object reach its maximum height? ty, max =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

1 - 1.) Projectile: At t = 0, an object is projected with a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online