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. The ball leaves his hand at a height of 2 feet with an initial velocity of 40 ft/sec. Use this information to answer the following questions. Be sure to show all the steps of your work. (Part 1) write a function representing the vertical position of the bal, s(t), in feet per second in terms of the time, t, after the ball leaves the child's hand. Use the position function S(t)= -1/2gf^2+v0t+sO. (Part 2) Using your results from part 1, determine the time interval for which the ball will be 18 feet high, where the child's friend in the tree can catch it. Show all your. Don't round your solutions! (Part 3). What is the maximum height the ball will reach and at what time will it reach that height? Show all your work.

Explanation / Answer

s(t) = so + ut - gt^2/2

so = intial height ; u = intial velocity ; g = 32.2 ft/sec^2

s(t) = 2 + 40t - 16.1t^2

When ball reaches 18ft ; s(t) =18

18 = 2 + 40t - 16.1t^2 ; 16.1 = 40t - 16t^2

4 = 10t - 4t^2 .solve for t

t = 0.504 sec and 2 sec

ball would reach maximum height at vertex : t = -b/2a = - (40/(2*-16.1) = 1.24 sec

Smax(1.2) = 2 + 40*1.24 - 16.1(1.24)^2 = 26.84 ft

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