A firecracker is shot upward off a balcony that is 10 feet above the ground with an initial velocity of v feet per second. The firecracker reaches a maximum height of 30.25 feet at 1.125 seconds after being shot upward. The formula for the vertical motion of an object is h(t) = \frac{1}{2} at {}^{2} + vt + sh(t)=21at2+vt+s
where the gravitational constant, a, is -32 feet per second, and h(t) is the height in feet modeled as a function of time, t.

The initial velocity of the firecracker is
A. 12
B. 27
C. 36
feet per second

The time when the firecracker reaches the ground is
A. 1.25
B. 2
C. 2.5
seconds

The correct answers are C and C I just need to understand why.
Thanks!

the answer is simply just 6n add the two together as they have an n

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step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

step-by-step explanation: