Question

A baseball is thrown in a parabolic arc. Help?

Answer 1

Answer:

The maximum height of the baseball is 9 feet

Step-by-step explanation:

we have

$p(t)=\frac{1}{2}gt^2+v_0t+p_0$

where

p(t) ----> baseball position above the ground in feet

t ----> the time in seconds

v_0 ----> is the initial velocity in ft/sec

p_0 ---> initial position above the ground

we have

$g=-32\frac{ft}{sec^2} \\\\v_0=16\frac{ft}{sec}\\\\p_0=5\ ft$

substitute the given values

$p(t)=\frac{1}{2}(-32)t^2+16t+5$

$p(t)=-16t^2+16t+5$

This is the equation of a vertical parabola open downward

The vertex represent a maximum

Convert the quadratic equation in vertex form

$p(t)=-16t^2+16t+5$

Factor -16 leading coefficient

$p(t)=-16(t^2-t)+5$

Complete the square

$p(t)=-16(t^2-t+\frac{1}{4})+5+4$

$p(t)=-16(t^2-t+\frac{1}{4})+9$

Rewrite as perfect squares

$p(t)=-16(t-\frac{1}{2})^2+9$

The vertex is the point (0.5,9)

The maximum height of the baseball above the ground is the y-coordinate of the vertex

therefore

The maximum height of the baseball is 9 feet