Question
Asked Sep 12, 2019
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Suppose that the height s of a ball (in feet) at time t (in seconds) is given by the formula s(t)=64−16(t−1)^2. This function is graphed below on the interval 0≤t≤3.

When (in seconds after being released) did the ball reach its highest point?

How long (in seconds) after being released did the ball hit the ground?

 

s(t
64
C
в
D
A
E
32
F
Gt
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s(t 64 C в D A E 32 F Gt

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Expert Answer

Step 1

Given a graph that represent the path of a ball. Height is defined with the help of function s(t) at time t is :

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s(f) 64-16(- 0sts3

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Step 2

First part:

Calculate the time(in seconds) when ball reach the highest point.

Here path of the ball is like a downward parabola.

Maximum height of the ball is defined in terms of the vertex of the parabola. SO, we will find the vertex of the parabola.

Step 3

In order to find the vertex of the parabola, we write the given function in form of vertex form of parabola.

The standard vertex form of parabola f(x) is : ...

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f(x) a(x-hk s)-16(-164 h 1 k 64 a 16

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