Afirework is launched into the air from ground level with an initial velocity of 128 ft/s. if acceleration due to gravity is –16 ft/s2, what is the maximum height reached by the firework?

[ h(t) = at2 + vt + h0 ]

a. 256 ft
b. 448 ft
c. 512 ft
d. 1,024 ft

Given that the equation to find the height of the firework is

h(t) = at² + vt + h₀

with a = -16 ft/s² and v = 128 ft/s. In addition, since the firework starts from the ground, then the initial height, h₀, is equal to 0. Substituting these values, we have

h(t) = -16t² + 128t + 0 = -16t² + 128t

Seeing that h(t) is a quadratic function, then it forms a parabola. To find its maximum height, we can compute for the parabola's vertex. 

To find the vertex's x-coordinate, we can use 

t = -b/2a = (-128)/(2 · -16) = -128/-32 = 4

Since, it takes 4 seconds for the firework to reach its maximum height, then the maximum height it reaches is equal to h(4). Hence, we have

h(4) = -16(4)² + 128(4) = -16(16) + 512 = 256

Hence, the highest that the firework can reach is equal to 256 ft. 

A. 256 ft

Correct answer A.

Step-by-step explanation:

2v

Step-by-step explanation:

Let the distance be d and average speed be v.

Half of the distance the speed was 2/3v

Let the speed be xv in the second half of the distance.

Then time is:

d/v = d/2 : (2/3v) + d/2 : (xv)d/v = 3/4(d/v) + (1/2x)(d/v)1 = 3/4 + 1/2x1/2x = 1 - 3/4 1/2x = 1/42x = 4x = 2

The rest of the road car has to ride with speed of 2v