218k views
2 votes
Can someone explain me problem #5 please? Thanks

Can someone explain me problem #5 please? Thanks-example-1

1 Answer

2 votes

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.

Given

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

Find

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.

Solution

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.

User Dxuhuang
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories