If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s

Question

If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s, its height (in meters) after t seconds is h = 13t − 0.83t².

 (Round your answers to two decimal places.)

(a) What is the velocity of the stone after 7 s?

(b) What is the velocity of the stone after it has risen 33 m?

Answer 1

Given: v = 13 m/s

h = 13t - 0.83t²

13t is the rock on the way up, and -0.83t² is the rock on the way down. ≅Note that as soon as the rock is launched, gravity is working against it, but on the way down, we have acceleration due to gravity.

a) v = ? after t = 7 s

Our original equation given to us was h = 13t - 0.83t².

1. Instantaneous velocity is the first derivative of position with respect to time, so, v = dh/dt.

Taking the derivative we get v = dh/dt = 13 - 1.66t.

Plugging in t = 7 s we get v = 1.38 m/s

b) v = ? after h = 33 m

Preliminaries:

This problem involves two time periods (t₁ = up, t₂ = down).

First we will solve for t_1∧2.

For h = 33 m and v = 13 m/s, h = 13t - 0.83t² becomes 33 = 13t - 0.83t².

We move the 13 over to the other side, set the equation equal to zero and then solve for the positive roots of t. This is messy, so remember the quadratic formula: -b ± √(b²- 4ac) / 2a.

In this case -0.83t² + 13t - 33 = 0 and to make it easier, we will multiply everything by 100 on both sides, that way decimals aren't an issue. So, we end up with: - 83t² + 1300t - 3300 = 0 where a = - 83, b = 1300, and c = - 3300.

So, we end up with: t_{1, 2} = - 1300 ± √(1300^2 - 4*(- 83)*(- 3300)) / 2*(- 83) ⇒

t↑ = 10(65 + √1486) / 83 ≈ 3.19 s (i)

t↓ = 10(65 - √1486) / 83 ≈ 12.48 s (ii)

So, we have t when h = 33 m both on the way up and on the way down. Since we are looking for the velocity when h = 33, we will use (i) and the formula we derived in part (a).

v = 13 - 1.66t = 13 - 1.66(3.19 s) = 7.70 m/s

Answer 2

velocity is the derivative of distance with respect to time so, in this case, dh/dt

Since h = 13t - 0.83t²

then dh/dt = 13 - (2)(0.83)t = 13 - 1.66t

so v(7) = 13 - 1.66(7) = 13 - 11.62 = 1.38 m/s

For b) we need to determine the time when h = 33 so:

33 = 13t - 0.83t²

0 = -0.83t² + 13t - 33

Using the quadratic formula:

t = 3.187 seconds or t = 12.476 seconds

The 3.187 seconds would be when the stone is going up and the 12.476 seconds would be when the stone is falling back to the moon.

v(3.187) = 13 - 1.66(3.187) = 13 - 5.29042 = 7.71 m/s