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Question

a. 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 ft with and 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 time, t, after the ball leaves the child's hand. Use the position function s(t)=-1/2gt^2 +v₀t +s_0.

b.Determine the time interval will be more than 18 ft high.

c.What is the maximum height the ball will reach and at what time will it reach that height? Write sentence to explain answer.

Answer 1

First we'll define our terms:

g is the acceleration due to gravity:   32 ft/sec²; t is time in seconds, where we define t=0 to be the time the ball leaves the child's hand; v₀ is initial velocity (given as +40 ft/sec); and s₀ is initial position (given as +2 ft).

 

Notes on a few subtleties:  signs:  we define positive distance as being above ground (ground s=0) and positive velocity as moving upward (the 1/2 gt² term is negative because gravity's acceleration is downward); dimensions: s(t) is one-dimensional, i.e., the height of the ball.  We are not concerned with position in the other two dimensions at all.  relativity: we may define t=0 to be any time we want and we may define s=0 to be any height we want; the ones we chose are simply the most convenient.

 

So, for part a):  s(t) = -1/2(32)t² + (40)t + 2

 

b) s(t) = 18 when -16t² + 40t + 2 = 18.

              Or, -16t² + 40t - 16 = 0.

divide by 8:  -2t² + 5t - 2 = 0

this factors to (-2t + 1)(t - 2) = 0.  So t = 2 or t = +1/2.  Then, between t = 1/2 and t = 2, the ball is over 18 feet high.

 

c) the ball reaches its maximum halfway between t = 1/2 and t = 2 which is t = 5/4.  At this time, the ball's height is 

          -1/2(32)(5/4)² + 40(5/4) + 2 = -16(25/16) + 10(5) + 2 = -25 + 50 + 2 = 27 ft.

  

   Alternatively, if you know calculus, the maximum of a function occurs when the derivative of the function equals zero.

  The derivative of s(t) is -32t + 40.  -32t + 40 = 0 when t = 40/32 = 5/4.