Question

A person standing cloes to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function
s(t)=-16t^2+64t+200
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
A) After how many seconds does the ball reach it's maximum height? What is the maximum height?
B) How many seconds does it take until the ball finally hits the ground?

Answer 1

Answer:

See below in bold.

Step-by-step explanation:

A).  s(t)= -16t^2+64t+200

Converting to vertex form.

s(t) = -16 (t^2 - 4t) + 200

s(t) =  -16 [(t - 2)^2 - 4] + 200

s(t) = -16(t - 2)^2 + 64 + 200

s(t) =  -16(t - 2)^2 + 264

The maximum height is after 2 seconds and it is 264 feet.

B) The ball hits the ground when s(t) = 0 so we have:

-16t^2+64t+200 = 0

x [-64 +/- sqrt (64^2 - 4*-16*200) ] / -32

x = 6.06, -2.06

So the answer is 6.06 seconds.