Question

acceleration and gravity, part 2
the distance covered by a freely falling body is calculated by the following formula,
$d = \frac{at^{2}}{2}$
where $d =$ distance
$a =$ acceleration
$t =$ time
example 1: how far will an object fall in 5 seconds?
answer: $d = \frac{(9.8 m/s^{2})(5s)^{2}}{2} = 122.5$ meters
example 2: what is the average velocity of a ball that attains a velocity of 39.2 m/s after 4 seconds?
answer: $v_{a}=\frac{v_{f}-v_{i}}{2}=39.2 - 0 = 19.6 m/s$
provide the answers to the questions below.

1. if a ball falls for 10 seconds, how far will it have traveled?

answer:

1. what is the final velocity of the ball after falling for 10 seconds?

answer:

1. what is the average velocity of the ball after falling for 10 seconds.

answer:

1. if a ball falls for 20 seconds, how far will it have traveled?

answer:

1. how long will it take an object dropped out a window to fall 176.4 meters?

answer:

1. if a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters, how long will it take to reach the ground?

answer:

1. what is the final vertical velocity of the box of supplies when it hits the ground?

answer:

1. what is the average vertical velocity of the box of supplies when it hits the ground?

answer:

Explanation:

Step1: Substitute values for question 1

Given \(a = 9.8\ m/s^{2}\), \(t = 10\ s\) into \(d=\frac{at^{2}}{2}\).
\(d=\frac{9.8\times10^{2}}{2}=\frac{9.8\times100}{2}\)

Step2: Calculate for question 1

\(d = 490\ m\)

Step3: Use \(v = at\) for question 2

Given \(a = 9.8\ m/s^{2}\), \(t = 10\ s\)
\(v=9.8\times10 = 98\ m/s\)

Step4: Use \(v_{a}=\frac{v_{f}-v_{i}}{2}\) for question 3

\(v_{i} = 0\), \(v_{f}=98\ m/s\)
\(v_{a}=\frac{98 - 0}{2}=49\ m/s\)

Step5: Substitute values for question 4

Given \(a = 9.8\ m/s^{2}\), \(t = 20\ s\) into \(d=\frac{at^{2}}{2}\)
\(d=\frac{9.8\times20^{2}}{2}=\frac{9.8\times400}{2}\)

Step6: Calculate for question 4

\(d = 1960\ m\)

Step7: Rearrange \(d=\frac{at^{2}}{2}\) for question 5

\(t=\sqrt{\frac{2d}{a}}\), given \(d = 176.4\ m\), \(a = 9.8\ m/s^{2}\)
\(t=\sqrt{\frac{2\times176.4}{9.8}}=\sqrt{\frac{352.8}{9.8}}=\sqrt{36}\)

Step8: Calculate for question 5

\(t = 6\ s\)

Step9: Rearrange \(d=\frac{at^{2}}{2}\) for question 6

\(t=\sqrt{\frac{2d}{a}}\), given \(d = 4410\ m\), \(a = 9.8\ m/s^{2}\)
\(t=\sqrt{\frac{2\times4410}{9.8}}=\sqrt{\frac{8820}{9.8}}=\sqrt{900}\)

Step10: Calculate for question 6

\(t = 30\ s\)

Step11: Use \(v = at\) for question 7

Given \(a = 9.8\ m/s^{2}\), \(t = 30\ s\) (from question 6)
\(v=9.8\times30=294\ m/s\)

Step12: Use \(v_{a}=\frac{v_{f}-v_{i}}{2}\) for question 8

\(v_{i} = 0\), \(v_{f}=294\ m/s\) (from question 7)
\(v_{a}=\frac{294 - 0}{2}=147\ m/s\)

Answer:

1. \(490\ m\)
2. \(98\ m/s\)
3. \(49\ m/s\)
4. \(1960\ m\)
5. \(6\ s\)
6. \(30\ s\)
7. \(294\ m/s\)
8. \(147\ m/s\)