This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 EAS 208 (Fall 2010) Homework #2 Solutions Problem 1381 The crossbar of the goalpost in American football is 10 c y ft = above the ground. To kick a field goal, the kicker must make the ball go between the two uprights supporting the crossbar, and the ball must be above the crossbar when it does so. Suppose that the kicker attempts a 40yd field goal ( 120 c x ft = ) and kicks the ball with initial velocity 70 o ft v s = and angle 40 o o = . By what vertical distance does the ball clear the crossbar? Answer: We choose our Cartesian coordinate system with xdirection the horizontal direction (positive to the right) and ydirection the vertical direction (positive upwards). The origin of the coordinate system is the point at the ground level where the ball is kicked. The initial velocity of the ball is resolved into its components in the x and y direction: cos40 o x V V = sin 40 o y V V = The acceleration of the ball in each direction is: x = && y g = && (acceleration due to gravity; assume constant) z = && , z = & , z = && (planar motion  xy plane) 2 Examine motion in each direction separately. For motion in the xdirection: x = && 1 1 1 0 x x dt C x dt C x C = + = + = & && & & ( ) 2 1 2 1 2 x x dt C x C dt C x C t C = + = + = + & Invoke initial conditions: ( ) 1 cos40 o x C V = = & and ( ) 2 x C = = Thus: x = && cos40 o x V = & cos40 o x Vt = For motion in the ydirection: y g = && ( ) 3 3 3 y y dt C y g dt C y gt C = + = + = + & && & & ( ) 2 4 3 4 3 4 1 2 y y dt C y gt C dt C y gt C t C = + = + + = + + & Invoke initial conditions: ( ) 3 sin 40 o y C V = = & and ( ) 4 y C = = Thus: y g = && sin 40 o y gt V = + & 2 1 sin 40 2 o y gt Vt = + 3 Therefore the solution for the distance traveled by the ball is: ( ) cos40 o x t Vt = ( ) 2 1 sin 40 2 o y t gt Vt = + Assume that the ball is over the crossbar at time c t . The position of the ball at that time according to our Cartesian coordinate system is ( ) , c cb x y , thus: ( ) 120 cos40 o c c c x t x Vt = = = ( ) 2 1 sin 40 2 o c cb c c y t y gt Vt = = + From the ( ) c x t equation we solve for time c t : ( ) 120 120 120 cos40 2.238 cos40 70cos40 o c c c c c c o o x t x Vt t t t s V = = = = = = Then substitute for the above result for time c t in the ( ) c y t equation: ( ) 2 2 2 2 1 1 sin 40 32.2 2.238 70 2.238 sin 40 2 2 o o c cb c c cb ft ft y t y gt Vt y s s s s = = + = + 20.06 cb y ft = The crossbar is 10 c y ft = above the ground, thus the ball clears the crossbar by (vertical distance): 20.06 10 10.06 clearance cb c clearance clearance y y y y ft ft y ft = = = 4 Problem 1386 At t = , a steel ball in a tank of oil is given a horizontal velocity 2 m s = v i . The components of the balls acceleration in 2 m s , are 1.2 x x a v = , 8 1.2 y y a v = , and 1.2 z z a v = . What is the velocity of the ball at 1 t s =...
View
Full
Document
This document was uploaded on 10/27/2011 for the course EAS 208 at SUNY Buffalo.
 Fall '08
 MehdiAhmadizadeh

Click to edit the document details