Pre-Calc Homework Solutions 418

Pre-Calc Homework Solutions 418 - 418 Section 10.4 (d)...

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

View Full Document Right Arrow Icon
27. (a) (Assuming that “ x ” is zero at the point of impact.) r ( t ) 5 ( x ( t )) i 1 ( y ( t )) j , where x ( t ) 5 (35 cos 27 8 ) t and y ( t ) 5 4 1 (35 sin 27 8 ) t 2 16 t 2 . (b) y max 5 } ( v 0 s 2 in g a ) 2 }1 4 5 } (35 si 6 n 4 27°) 2 }1 4 < 7.945 feet, which is reached at t 5 } v 0 s g in a } = } 35 s 3 in 2 27° } < 0.497 seconds. (c) For the time, solve y 5 4 1 (35 sin 27°) t 2 16 t 2 5 0 for t , using the quadratic formula: t 5 < 1.201 seconds. Then the range is about x (1.201) 5 (35 cos 27°)(1.201) < 37.406 feet. (d) For the time, solve y 5 4 1 (35 sin 27°) t 2 16 t 2 5 7 for t , using the quadratic formula: t 5 < 0.254 and 0.740 seconds. At those times the ball is about x (0.254) 5 (35 cos 27°)(0.254) < 7.906 feet and x (0.740) 5 (35 cos 27°)(0.740) < 23.064 feet from the impact point, or about 37.460 2 7.906 < 29.554 feet and 37.460 2 23.064 < 14.396 feet from the landing spot. (e)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online