Lesson A tangled problem on a ball thrown upward

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A tangled problem on a ball thrown upward


Problem 1

At time  t= 0,  a ball was thrown upward.
The ball's height,  in feet,  after  t  seconds is given by the function   h(t) = c+-+%28d-4t%29%5E2,
in which both  c  and  d  are positive constants.  If the ball reached its maximum height
of  104  feet at time  t= 3  seconds,  what is the height,  in feet,  of the ball at time  t= 1.5 seconds ?

Solution

In this problem, the standard quadratic function of height, h(t) = -16t^2 + ut + h0,  is presented

in vertex form, and your first task is to identify the parameters of this vertex form.



    I will help you to identify these parameters.



First, the term "c" represents the highest position of the ball, which is given in the problem as 104 feet.

So, c = 104 feet.



Second, d - 4t = 0  determines the time "t", when the highest position is reached.

The problem says that the maximum height is reached at t= 3 seconds;  hence, d= 4t = 4*3 = 12 seconds.



    Now you know everything about your function: it is  h(t) = 104 - (12-4t)^2.



Now to answer the problem's question, you simply substitute t= 1.5 seconds in the last formula.

You get then


    h(1.5) = 104 - (12 - 4*1.5)^2 = 68 feet.    ANSWER


Thus, the solution of this tangled problem is completed.


My other lessons in this site on a projectile thrown/shot/launched vertically up, or horizontally, or at an angle to horizon
    - Introductory lesson on a projectile thrown-shot-launched vertically up
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform
    - A flare is launched from a life raft vertically up
    - A soccer ball - write the height equation in vertex form
    - OVERVIEW of lessons on a projectile thrown/shot/launched vertically up

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