CHAPTER 3 - 14 - 0 = ( v 2 sin ) 2 + 2( g ) h max , which...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 = ( v 2 sin ) 2 + 2( g ) h max , which gives h max = v 2 sin 2 /2 g . When we equate this to the range, we get v 2 sin 2 /2 g = 2 v 2 sin cos / g , which gives tan = 4, = 76.0 . 43. The ball passes the goal posts when it has traveled the horizontal distance of 36.0 m. From this we can find the time when it passes the goal posts: x = v x t ; 36.0 m = (20.0 m/s) cos 37.0 t , which gives t = 2.25 s. To see if the kick is successful, we must find the height of the ball at this time: y = y + v y t + ! a y t 2 = 0 + (20.0 m/s) sin 37.0 (2.25 s) + ! ( 9.80 m/s 2 )(2.25 s) 2 = 2.24 m. Thus the kick is unsuccessful because it passes 0.76 m below the bar....
View Full Document

CHAPTER 3 - 14 - 0 = ( v 2 sin ) 2 + 2( g ) h max , which...

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

View Full Document
Ask a homework question - tutors are online