Question

Suppose that the height s of a ball (in feet) at time t (in seconds) is given by the formula s(t) = 64 - 16(t - 1)^2. This function is graphed below on the interval 0 ≤ t ≤ 3. Which labeled point (A-G) corresponds to the release of the ball? How high (in feet) was the ball when it was released? Which labeled point corresponds to the highest point of the ball? When (in seconds after being released) did the ball reach its highest point? Which labeled point corresponds to the ball hitting the ground? How long (in seconds) after being released did the ball hit the ground?

Answer 1

Answer:

Kindly check explanation

Step-by-step explanation:

Suppose that the height s of a ball (in feet) at time t (in seconds) is given by the formula

s(t) = 64 - 16(t - 1)^2

t interval = 0 ≤ t ≤ 3

1) point A (from the graph)

11) Height of ball when it was released

Ball was released at t = 0

s(0) = 64 - 16(0 - 1)^2

= 64 - 16(-1)^2

= 64 - 16(1)

= 64 - 16

= 48 feets

111) point C ( from the graph)

IV) highest point of the ball is 64

Hence,

s(t) = 64 - 16(t - 1)^2

64= 64 - 16(t - 1)^2

16(t - 1)^2 = 64 - 64

16(t - 1)^2 = 0

16t^2 - 32t + 16 = 0

t^2 - 2t + 1 = 0

(t-1) = 0 (t-1) = 0

t = 1

V) Point G (from graph)

V1)

height = 0

s(t) = 64 - 16(t - 1)^2

0 = 64 - 16(t - 1)^2

16(t - 1)^2 = 64

(t - 1)^2 = 64/16

(t - 1)^2 = 4

(t - 1)² = 2²

t - 1 = 2

t = 2 + 1

t = 3