{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

excredsolns

# excredsolns - s = v t ½ at 2 = 0 0.5(3 m/s 2(4.5 s 2 =...

This preview shows page 1. Sign up to view the full content.

PHYS 15200 Extra Credit Assignment - SOLUTION A cyclist starts at rest at the top of a steep 68 ° incline of height 50 m. The cyclist accelerates down the incline at a constant 3 m/s 2 . Exactly 4.5 s later, he throws a ball at 10 m/s perpendicularly away from the surface of the incline (relative to him). (A) What is the initial velocity of the ball relative to the ground? (B) Where does the ball strike the ground? At 4.5 s, the cyclist’s velocity is v = v 0 + at = 0 + (3 m/s 3 )(4.5 s) = 13.5 m/s and he has traveled
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s = v t + ½ at 2 = 0 + 0.5(3 m/s 2 )(4.5 s) 2 = 30.375 m down the incline. (A) v = 13.5 2 + 10 2 = 16.8 m/s θ = tan-1 (10/13.5) = 36.5 ° Answer : 16.8 m/s at -31.5 ° (B) 30.375 m First find time to hit ground: 28.163 m Set y = y + v y t − ½ gt 2 = 0 t = 1.40 s v x = 16.8cos31.5 ° v y = − 16.8sin31.5 ° 21.837 m 11.379 m 8.82 m x = v x t = (16.8cos31.5 ° )(1.40 s) = 20.0 m Answer : 11.2 from the base of the incline 50 m 68 ° 22 ° θ 13.5 m/s 10 m/s 68 ° y 0...
View Full Document

{[ snackBarMessage ]}