FINAL STAGE 2022
Presbyterian Boys' Secondary School (PRESEC), Legon: 50 points
Prempeh College: 41 points
Adisadel College: 32 points
ROUND 2 - SPEED RACE
QUESTION
An object is dropped from rest 196 meters above ground. 2.0 seconds later, another object is dropped from the same height so that both hit the ground at the same time.
Find the initial speed of the second object.
ANSWER: $\approx24.1$ m/s
SOLUTION:
Find the fall time of the first object (dropped from rest):
$196=\dfrac{1}{2}gt_1^2$
$t_1=\sqrt{\dfrac{2(196)}{9.8}}$
$t_1=\sqrt{40}$
$t_1\approx6.32\text{ s}$
The second object has less time to cover the same 196 m, since it starts 2.0 s later:
$t_2=t_1-2.0$
$t_2\approx4.32\text{ s}$
It is launched downward with initial speed $u$:
$196=ut_2+\dfrac{1}{2}gt_2^2$
$u=\dfrac{196-\dfrac{1}{2}(9.8)(4.32)^2}{4.32}$
$u\approx24.1\text{ m/s}$
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PRACTICE QUESTIONS
1. QUESTION: An object is dropped from rest 19.6 m above the ground. 1.0 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 15 m/s or $1.5\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{19.6}{4.9}$
$t_1^2 = \dfrac{196}{49}$
$t_1^2 = 4.00\text{ s²}$
$t_1 = 2.00\text{ s}$
$t_2 = t_1 - 1.0$
$t_2 = 1.0\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 1.0^2$
$\dfrac{1}{2}gt_2^2 = 4.9\text{ m}$
$u = \dfrac{19.6 - 4.9}{1.0}$
$u = \dfrac{14.7}{1.0}$
$u = 147\times10^{-1}$
$u \approx 15\text{ m/s}$
2. QUESTION: An object is dropped from rest 44.1 m above the ground. 2.0 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 39 m/s or $3.9\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{44.1}{4.9}$
$t_1^2 = \dfrac{441}{49}$
$t_1^2 = 9.00\text{ s²}$
$t_1 = 3.00\text{ s}$
$t_2 = t_1 - 2.0$
$t_2 = 1.0\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 1.0^2$
$\dfrac{1}{2}gt_2^2 = 4.9\text{ m}$
$u = \dfrac{44.1 - 4.9}{1.0}$
$u = \dfrac{39.2}{1.0}$
$u = 392\times10^{-1}$
$u \approx 39\text{ m/s}$
3. QUESTION: An object is dropped from rest 78.4 m above the ground. 2.0 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 29 m/s or $2.9\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{78.4}{4.9}$
$t_1^2 = \dfrac{784}{49}$
$t_1^2 = 16.0\text{ s²}$
$t_1 = 4.00\text{ s}$
$t_2 = t_1 - 2.0$
$t_2 = 2.0\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 2.0^2$
$\dfrac{1}{2}gt_2^2 = 19.6\text{ m}$
$u = \dfrac{78.4 - 19.6}{2.0}$
$u = \dfrac{58.8}{2.0}$
$u = \dfrac{588\times10^{-1}}{2}$
$u \approx 29\text{ m/s}$
4. QUESTION: An object is dropped from rest 78.4 m above the ground. 3.00 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 73.5 m/s or $7.35\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{78.4}{4.9}$
$t_1^2 = \dfrac{784}{49}$
$t_1^2 = 16.0\text{ s²}$
$t_1 = 4.00\text{ s}$
$t_2 = t_1 - 3.00$
$t_2 = 1.00\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 1.00^2$
$\dfrac{1}{2}gt_2^2 = 4.9\text{ m}$
$u = \dfrac{78.4 - 4.9}{1.00}$
$u = \dfrac{73.5}{1.00}$
$u = 735\times10^{-1}$
$u = 73.5\text{ m/s}$
5. QUESTION: An object is dropped from rest 176.4 m above the ground. 2.00 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 24.5 m/s or $2.45\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{176.4}{4.9}$
$t_1^2 = \dfrac{1764}{49}$
$t_1^2 = 36.0\text{ s²}$
$t_1 = 6.00\text{ s}$
$t_2 = t_1 - 2.00$
$t_2 = 4.00\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 4.00^2$
$\dfrac{1}{2}gt_2^2 = 78.4\text{ m}$
$u = \dfrac{176.4 - 78.4}{4.00}$
$u = \dfrac{98}{4.00}$
$u = \dfrac{49}{2}$
$u = 24.5\text{ m/s}$
6. QUESTION: An object is dropped from rest 176.4 m above the ground. 3.00 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 44.1 m/s or $4.41\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{176.4}{4.9}$
$t_1^2 = \dfrac{1764}{49}$
$t_1^2 = 36.0\text{ s²}$
$t_1 = 6.00\text{ s}$
$t_2 = t_1 - 3.00$
$t_2 = 3.00\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 3.00^2$
$\dfrac{1}{2}gt_2^2 = 44.1\text{ m}$
$u = \dfrac{176.4 - 44.1}{3.00}$
$u = \dfrac{132.3}{3.00}$
$u = \dfrac{1323\times10^{-1}}{3}$
$u = 44.1\text{ m/s}$
7. QUESTION: An object is dropped from rest 176.4 m above the ground. 4.00 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 78.4 m/s or $7.84\times10^{1}$ m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{176.4}{4.9}$
$t_1^2 = \dfrac{1764}{49}$
$t_1^2 = 36.0\text{ s²}$
$t_1 = 6.00\text{ s}$
$t_2 = t_1 - 4.00$
$t_2 = 2.00\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 2.00^2$
$\dfrac{1}{2}gt_2^2 = 19.6\text{ m}$
$u = \dfrac{176.4 - 19.6}{2.00}$
$u = \dfrac{156.8}{2.00}$
$u = \dfrac{1568\times10^{-1}}{2}$
$u = 78.4\text{ m/s}$
8. QUESTION: An object is dropped from rest 4.90 m above the ground. 0.500 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 7.35 m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{4.90}{4.9}$
$t_1^2 = \dfrac{49}{49}$
$t_1^2 = 1.00\text{ s²}$
$t_1 = 1.00\text{ s}$
$t_2 = t_1 - 0.500$
$t_2 = 0.500\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 0.500^2$
$\dfrac{1}{2}gt_2^2 = 1.225\text{ m}$
$u = \dfrac{4.90 - 1.225}{0.500}$
$u = \dfrac{3.675}{0.500}$
$u = \dfrac{3675\times10^{-3}}{5\times10^{-1}}$
$u = 7.35\text{ m/s}$
9. QUESTION: An object is dropped from rest 44.1 m above the ground. 0.500 s later, another object is thrown straight down from the same height so that both hit the ground at the same time. Find the initial speed of the second object.
ANSWER: 5.39 m/s
SOLUTION:
$h = \dfrac{1}{2}gt_1^2$
$t_1^2 = \dfrac{2h}{g}$
$t_1^2 = \dfrac{h}{4.9}$
$t_1^2 = \dfrac{44.1}{4.9}$
$t_1^2 = \dfrac{441}{49}$
$t_1^2 = 9.00\text{ s²}$
$t_1 = 3.00\text{ s}$
$t_2 = t_1 - 0.500$
$t_2 = 2.50\text{ s}$
$h = ut_2 + \dfrac{1}{2}gt_2^2$
$u = \dfrac{h - \dfrac{1}{2}gt_2^2}{t_2}$
$\dfrac{1}{2}gt_2^2 = 4.9 \times 2.50^2$
$\dfrac{1}{2}gt_2^2 = 30.625\text{ m}$
$u = \dfrac{44.1 - 30.625}{2.50}$
$u = \dfrac{13.475}{2.50}$
$u = \dfrac{13475\times10^{-3}}{25\times10^{-1}}$
$u = 5.39\text{ m/s}$