A child throws a ball straight upwards to his friend who is sitting in a tree 18

Question

A child throws a ball straight upwards to his friend who is sitting in a tree 18 ft above ground level. If the ball leaves his hand at a height of 2 feet with an initial velocity of 40 ft/sec, 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 s (t)=-1/2gt^2+v0t+s0. (The zero's are subscripts.) Be sure to simplify the first term!

Using your result from part 1, determine the time interval for which the ball will be more than 18ft high, where the child's friend in the tree can catch it. Write you final answer in both compound inequality form and a sentence.

What is the maximum height the ball will reach and at what time will it reach that height?

Explanation / Answer

a) s(t) = -gt^2/2 +vo*t +so

we have : vo = intial velocity = 40 ft/se

so = height from which ball is thown = 2 feet

constant g = 32.3 ft/sec^2

synbstitute these values:

s(t) = -32.2t^2/2 + 40t +2

= -16.1t^2 +40t +2

b) s(t) > 18 ft

18 >-16.1t^2 +40t +2

-16.1t^2 +40t -16 <0

solve the inequality: roots of -16.1t^2 +40t -16 =0

t= 0.5 and t=1.98

inequlaity solution : t< 0.5 and t > 1.98

t must be >0

So, 0< t < 0.5 and t>1.98 sec

So, fot time greatern than 0 and less than 0.5 sec, height of ball is gtreater than 18 ft

also, whent t> 1.98 sec height of ball is < 18 feet

Maximum height of ballat what time:

for a quadratice function max, height is attianed at vertex: t = -b/2a

t = -(40)/2(-16.1) = 1.24

maximum height s(1.24) = -16.1(1.24)^2 +40*1.24 +2 = 26.84 feet

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