Every year in Delaware there is a contest where people create cannons and catapults designed to launch pumpkins as far in the air as possible. The equation below can be used to represent the height, y, of a launched
pumpkin, where x is the time in seconds that the pumpkin has been in the
air. What is the maximum height that the pumpkin reaches? How many
seconds have passed when the pumpkin hits the ground? (Hint: If the
pumpkin hits the ground. its height is O feet.)

y = 15 + 110x – 16x^2

A-The pumpkin's maximum height is 204.06 feet and it hits the ground after 7.01
seconds.

B-The pumpkin's maximum height is 3.44 feet and it hits the ground after 7.01 seconds.

C-The pumpkin's maximum height is 204.06 feet and it hits the ground after 3.44
seconds.

D-The pumpkin's maximum height is 3.44 feet and it hits the ground after 204.06
seconds.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)