Please help A cannonball shot from the top of a building
can be modeled by the following equation:
h -16(t 3) 189
Where t is in seconds and h is in feet.

A) How high is the cannon? Consider how
this is the location of the cannon ball at
second 0. Show or explain your answer.

B)When is the cannonball at the highest point? show or explain your answer.

a. 45 ft

b. 3 seconds

Step-by-step explanation:

a.

At time equals zero, the cannon is fired. The cannonball at that time is just exiting the cannon, so the cannonball at time zero is at the same height as the cannon.

We let t = 0, and evaluate function h(t).

h(t) = -16(t - 3)^2 + 189

h(0) = -16(0 - 3)^2 + 189

h(0) = -16(9) + 189

h(0) = -144 + 189

h(0) = 45

The height of the cannon is 45 ft.

b.

The cannonball is fired. It starts at 45 ft height which is where the cannon is. It goes up until it reaches a maximum height. Then it starts falling to the ground. At some point after it reaches maximum height, it reaches a height of 45 ft as it falls to the ground. We can find the time it takes the cannonball to be at 45 ft height on the way down.

We set the function equal to 45 and solve for t. One answer will be 0 since we already know the height at time zero is 45 ft. We need to find the other time.

-16(t - 3)^2 + 189 = 45

-16(t - 3)^2 = -144

(t - 3)^2 = 9

t - 3 = 3 or t - 3 = -3

t = 6 or t = 0

The other time that the height of the cannonball is zero is 6 seconds.

The graph is symmetric from t = 0 to t = 6. The maximum height happens exactly in the middle of the time period of 0 to 6 seconds. The maximum height happens at t = 3 seconds.

cheater

step-by-step explanation:

answer: rolling a number cube and flipping a coin.

step-by-step explanation: