Question

A person standing close to the edge on top of a 28-foot building throws a ball vertically upward. The quadratic function h = − 16 t 2 + 108 t + 28 models the ball's height about the ground, h , in feet, t seconds after it was thrown. A) After how many seconds does the ball reach it's maximum height? What is the maximum height?B) How many seconds does it take until the ball finally hits the ground? C) Find s(0) and describe what this means?

Answer 1

Answer:

Part.A

t=3.345 s

Maximum height = h= 210.24 feet

Part.B

t=44 s

Part.C

S(0)=28 feet

Step-by-step explanation:

Given:

maximum height = 28 feet

quadratic equation for the height is h= − 16 t 2 + 108 t + 28

Part. A: Maximum height=? and Time for maximum height=?

As the given equation is equation of parabola and the maximum height is at the vertex of the graph for the equation of parabola.

For finding the vertex of parabola is we need it's x and y coordinates.

x-coordinate of vertex is the time for maximum for maximum height while y-coordinate is the maximum height.

formula for x-coordinate is -b/2a

comparing the given equation with the original equation of parabola we get the values of a, b and c as;

a=-16, b=108, and c=28

Here, x-coordinate is t=-b/2a

t=-108/(2*-16)

t=-108/(-32)

t=3.345 s

for y-coordinate (h), put the value of x-coordinate (i.e. value of t) in the given equation of parabola as;

h=-16(3.345)^2+108*3.345+28

h= -179.02+361.26+28

h=210.24 feet

Maximum height = h= 210.24 feet

Part.B How many seconds does it take until the ball finally hits the ground?

If ball hits the ground it means y-component is zero.

putting y=0 in the given equation we get the time that the ball will take to hit the ground

0=− 16 t^2 + 108 t + 28

using quadratic formula, we get

$t=-\frac{b}{2}$ ± $\frac{ \sqrt{b^{2}-4ac } }{2}$

t=$\frac{-(128)}{2}$± $\frac{\sqrt{108^{2}-4(-16)(28) } }{2}$

t=-14±$\frac{116}{2}$

t=-14+58 or -14-58

t=44 s or -72

time cannot be negative so t=44 s

Part.C Find s(0) and describe what this means?

given equation for h can also be denoted by s(t) which represents distance as a function of time (height is also a distance)

S(0)=-16*0^2+108*0+28

S(0)=28 feet

S(0) means the initial distance when ball was at the edge on the top of building at initial time i.e. when t=0 s