Can someone explain me problem #5 ?

The question describes the motion of a ball and asks about the motion of a rocket. There is no explanation for that. It goes directly into my "dumb question" file.

The equation of ballistic motion is

... h(t) = -1/2·g·t² + v₀·t + h₀

where v₀ and h₀ are the initial upward velocity and height, respectively.

A rocket's motion is modeled by

... h(t) = -4.9t² + 18t + 14 . . . . . where h is in meters and t is in seconds

Part A: v₀

Part B: h₀

Part C: (1) changes if h were measured in feet instead of meters. (2) the gravity coefficient in ft/s².

Parts A and B: You have correctly identified the coefficients in the equation that tell you the initial velocity (18 m/s) and height (14 m).

Part C: The relationship between feet and meters is

... 0.3048 m = 1 ft

Changing the units from m/s and m/s² to ft/s and ft/s² means we need to multiply the equation's coefficients by the appropriate conversion factor. Starting with the first coefficient ...

... (-1/2)g = -4.9 m/s²

... g = 9.8 m/s² = (9.8/0.3048) ft/s² ≈ 32.15 ft/s²

The new initial velocity is

... 18 m/s = 18/0.3048 ft/s ≈ 59.01 ft/s

The new initial height is

... 14 m = 14/0.3048 ft ≈ 45.93 ft

If we round all coefficients to 2 significant digits, the motion equation for units of feet and seconds is

... h(t) = -16t² + 59t + 46

(1) All of the coefficients in the function change to appropriate units. The new equation is ...

... h(t) = -16t² + 59t + 46

(2) The gravity coefficient is approximately 32 ft/s², so shows up in the equation as a coefficient of -16.

a) t'(-3, 6) and v'(0, 3)  

step-by-step explanation:

the answer would be  

a.

3: 1

step-by-step explanation: