Question

A child throws a ball straight upwards to his friend who is sitting in a tree 14 ft above ground level. The ball leaves his hand at a height of 3 feet with an initial velocity of 30 ft/sec. Use this information to answer the following questions. Be sure to show all steps of your work using the equation editor.

Part 1: (2 pts)

Write a function representing the vertical position of the ball, s(t), in feet per second in term of the time, t, after the ball leaves the child's hand. Use the position function . Refer to video 1.5.2 for a refresher of part 1. Be sure to simplify the first term!

Part 2: (6 pts)

Using your result from part 1, determine the time interval for which the ball will be 14 feet high, where the child's friend in the tree can catch it. Show all work using the equation editor. Write your final answer in sentence and interval form. Don't round your solutions!

Part 3: (4 pts)

What is the maximum height the ball will reach and at what time will it reach that height? Show work using the equation editor and write a sentence that explains your answer. Don't round your solutions.

Answer 1

initial height = 3 feet

velocity = 30 ft / sec

so, equation becomes

s(t) = - 1/2 ( 32) t^2 + 30t + 3

s(t) = - 16t^2 + 30t + 3

b)

plugging s(t) = 14

14 = - 16t^2 + 30t + 3

16t^2 - 30t + 11 = 0

t = [ 30 +- sqrt ( 30^2 - 4* 11 * 16 ) ] / 2(16)

t = 0.5 , 1.375

time interval for which ball will be 14 feet high is

[ 0.5 , 1.375 ]

c)

maximum height occurs at

t =- b/2a

t = - 30 / 2 ( - 16 )

t = 30/32 = 15/16

maximum height is

s(t) = - 16 (15/16)^2 + 30(15/16)+ 3 = 273/16

maximum height = 273/16 feet it occurs after 15/16 seconds