If a rock was thrown at t-2.23tp and hits its height at 1.56 meters what was the height of the tower?

t = 4.875 s

h(4.875) = 686.25 ft

t ≈ 11.424 s

Step-by-step explanation:

These questions are more easily answered if the equation is written in vertex form. Subtracting the constant and dividing out -16, we get ...

h(t) -306 = -16(t^2 -9.75t)

Adding the square of half the t-coefficient inside parentheses gives ...

 h(t) -306 -380.25 = -16(t^2 -9.75t +23.765625)  

 h(t) = -16(t -4.875)^2 + 686.25

This tells us the maximum height is 686.25 ft at time t = 4.875 seconds.

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When the rock hits the ground, h(t) = 0, so ...

0 = -16(t -4.875)^2 + 686.25

686.25/16 = (t -4.875)^2 . . . . . subtract 686.25, divide by -16

√42.890625 +4.875 = t ≈ 11.424 . . . . take the square root, add 4.875

The rock hits the ground after about 11.424 seconds.

it should be a and d. lmk if i'm wrong xd

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: