Sven shoots a cannon ball out of a cannon and the path of the ball is modeled in the graph. How long was the cannon ball in the air for (round to the nearest second?
Path of Cannonball
Height feet)
Time (seconds
a)8 secs
b) about 14 secs
c) about 15 secs
d) about 16 secs

d. 16 secs

Step-by-step explanation:

Each tick mark on the x-axis is 1 second. The cannon ball is in the air from when it is shot out to when it hits the ground so look for the positive x-intercept. It was in the air for about 16 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.

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