Question

A rocket is launched vertically from the ground with an initial velocity of 64. ft/sec Write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched. Graph on the coordinate plane. Use your graph from Part 3(b) to determine the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air. Answer:

Answer 1

Answer:

a. s = 64t - 16t²

a. maximum height = 64 ft

b. time it took to reach maximum height = 2 s

c. total time it spent in the air = 4 s

Step-by-step explanation:

a. Using Newton's third equation of motion under gravity, s = ut - 1/2gt² where s = vertical height of rocket, t = time, u = initial velocity of rocket = 64 ft/s and g = acceleration due to gravity = 32 ft/s².

So, s = 64t - 1/2(32)t² = 64t - 16t²

The equation is thus s = 64t - 16t²

b. The maximum height is obtained when ds/dt = 0

ds/dt = d[64t - 16t²]/dt = 64 - 32t

64 - 32t = 0

32t = 64

t = 64/32 = 2 s

Substituting t = 2 into s, we have

s = 64(2) - 16(2)² = 128 - 64 = 64 ft

c. The amount tot time it took to reach maximum height is 2 s

d. The amount of time it was in the air is obtained when s = 0

s = 0 ⇒

64t - 16t² = 0

16t(4 - t) = 0

16t = 0 or 4 - t = 0

t = 0 or t = 4

So the rocket spends 4 s in the air.

The values of the different variable for a, b,c and d are indicated in the graph.

The point (2,64) indicated the time for maximum height 2 s and maximum height 64 ft.

The point (4,0) indicates the total time spent in the air 4 s