9-10-ProjectileMotionII

# 9-10-ProjectileMotionII - Projectile Motion(General Launch...

This preview shows pages 1–3. Sign up to view the full content.

Projectile Motion (General Launch Angle) t x x 0x 0 v + = 2 0 2 1 t a t y y y + + = 0y v H The trajectory is a parabola a y = - 9.8 m/s 2 y 0 v Projectile Motion t v x v y x x 0 v v = t a y y y + = 0 v v x 0 v x 0 v x 0 v x 0 v x 0 v y 0 v y 0 v - 0 v

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

View Full Document
Example H s m s m v v o x / 32 . 15 40 cos ) / 20 ( cos 0 0 = = = θ s m s m v v o y / 86 . 12 40 sin ) / 20 ( sin 0 0 = = = To find Maximum height, go to y motion. v 0y = 12.86m/s At maximum height: v y = 0 a y = - 9.8m/s 2, Find y=? ) ( 2 0 2 0 2 y y a v v y y y = initial position x 0 = 0, y 0 = 0 y ) 8 . 9 ( 2 ) 86 . 12 ( 0 2 2 = m y 44 . 8 = maximum height H = 8.44m y v 0 ＝０ y v ＝０ y v 2. How much time does it take the ball to reach the maximum height? 1. Find the maximum height H the ball attains. A football is kicked at an angle of θ =40 o above the horizontal axis. Its initial speed is 20 m/s. Ignoring air resistance t ) 8 . 9 ( 86 . 12 0 s t 31 . 1 = t a v v y y y + = 0 Example (continued) 3. How long does the ball spend in the air? H
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/06/2010 for the course PHYSICS 1220 taught by Professor Yung during the Fall '09 term at Missouri (Mizzou).

### Page1 / 4

9-10-ProjectileMotionII - Projectile Motion(General Launch...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online