QUARTER FINAL STAGE 2024
Accra Academy: 45 points
Oyoko Methodist SHS: 36 points
St. Joseph Seminary SHS: 28 points
ROUND 1
FIRST SET OF QUESTIONS
PREAMBLE
An object is projected from ground level at $15^{\circ}$ to the horizontal.
$\text{Range} = \dfrac{u^2 \sin(2\theta)}{g}$
QUESTION 1
Find the maximum horizontal range for the object with an initial speed of $40\text{ m/s}$.
ANSWER: $80\text{ m}$
SOLUTION:
$\theta=15^{\circ}$, so $2\theta=30^{\circ}$ and $\sin(30^{\circ})=0.5$.
$\text{Range}=\dfrac{40^2\times0.5}{10}$
$\text{Range}=\dfrac{1600\times0.5}{10}$
$\text{Range} = \dfrac{1600 \times 5\times10^{-1}}{10}$
$\text{Range} = 160 \times 5 \times10^{-1}$
$\text{Range}=80\text{ m}$
QUESTION 2
Find the initial speed of the object required to achieve a horizontal range of $720\text{ m}$.
ANSWER: $120\text{ m/s}$
SOLUTION:
Rearrange for $u^2$:
$u^2=\dfrac{\text{Range}\times g}{\sin(2\theta)}$
$u^2=\dfrac{720\times10}{0.5}$
$u^2 = \dfrac{720 \times 10}{5\times10^{-1}}$
$u^2 = 144 \times 10 \times10^{1}$
$u^2=14400$
$u=\sqrt{14400}$
$u=120\text{ m/s}$
QUESTION 3
Find the horizontal range for the object with an initial speed of $20\text{ m/s}$.
ANSWER: $20\text{ m}$
SOLUTION:
$\text{Range}=\dfrac{20^2\times0.5}{10}$
$\text{Range}=\dfrac{400\times0.5}{10}$
$\text{Range} = \dfrac{400 \times 5\times10^{-1}}{10}$
$\text{Range} = 40 \times 5 \times10^{-1}$
$\text{Range}=20\text{ m}$
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PRACTICE QUESTIONS
1. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 14.0 m/s.
ANSWER: 10.0 m or $1.00\times10^{1}$ m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$R = \dfrac{14.0 \times 14.0 \times 0.500}{9.8}$
$R = \dfrac{14 \times 14 \times 5}{98}$
$R = \dfrac{14 \times 5}{7}$
$R = 2 \times 5$
$R = 10.0\text{ m}$
2. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 21.0 m/s.
ANSWER: 22.5 m or $2.25\times10^{1}$ m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$R = \dfrac{21.0 \times 21.0 \times 0.500}{9.8}$
$R = \dfrac{21 \times 21 \times 5}{98}$
$R = \dfrac{3 \times 21 \times 5}{14}$
$R = \dfrac{3 \times 3 \times 5}{2}$
$R = 22.5\text{ m}$
3. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 35.0 m/s.
ANSWER: 62.5 m or $6.25\times10^{1}$ m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$R = \dfrac{35.0 \times 35.0 \times 0.500}{9.8}$
$R = \dfrac{35 \times 35 \times 5}{98}$
$R = \dfrac{5 \times 35 \times 5}{14}$
$R = \dfrac{5 \times 5 \times 5}{2}$
$R = 62.5\text{ m}$
4. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 9.80 m/s.
ANSWER: 4.90 m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$R = \dfrac{9.80 \times 9.80 \times 0.500}{9.8}$
$R = \dfrac{98\times10^{-1} \times 98\times10^{-1} \times 5\times10^{-1}}{98\times10^{-1}}$
$R = 98 \times 5 \times10^{-2}$
$R = 4.90\text{ m}$
5. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 1.40 m/s.
ANSWER: 0.100 m or $1.00\times10^{-1}$ m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$R = \dfrac{1.40 \times 1.40 \times 0.500}{9.8}$
$R = \dfrac{14\times10^{-1} \times 14\times10^{-1} \times 5\times10^{-1}}{98\times10^{-1}}$
$R = \dfrac{14 \times 5}{7} \times10^{-2}$
$R = 2 \times 5 \times10^{-2}$
$R = 0.100\text{ m}$
6. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the horizontal range for an initial speed of 16 m/s.
ANSWER: 13 m or $1.3\times10^{1}$ m
SOLUTION:
$R = \dfrac{u^2\sin 2\theta}{g}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$g = 9.8$
$g \approx 10\text{ m/s}^2$
$R = \dfrac{16 \times 16 \times 0.500}{10}$
$R = \dfrac{16 \times 16 \times 5\times10^{-1}}{10}$
$R = \dfrac{16 \times 16}{2} \times10^{-1}$
$R = 8 \times 16 \times10^{-1}$
$R \approx 13\text{ m}$
7. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the initial speed needed for a horizontal range of 10.0 m.
ANSWER: 14.0 m/s or $1.40\times10^{1}$ m/s
SOLUTION:
$u^2 = \dfrac{Rg}{\sin 2\theta}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$u^2 = \dfrac{10.0 \times 9.8}{0.500}$
$u^2 = \dfrac{10 \times 98}{5}$
$u^2 = 2 \times 98$
$u^2 = 196\text{ m}^2/\text{s}^2$
$u = \sqrt{196}$
$u = 14.0\text{ m/s}$
8. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the initial speed needed for a horizontal range of 40.0 m.
ANSWER: 28.0 m/s or $2.80\times10^{1}$ m/s
SOLUTION:
$u^2 = \dfrac{Rg}{\sin 2\theta}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$u^2 = \dfrac{40.0 \times 9.8}{0.500}$
$u^2 = \dfrac{40 \times 98}{5}$
$u^2 = 8 \times 98$
$u^2 = 784\text{ m}^2/\text{s}^2$
$u = \sqrt{784}$
$u = 28.0\text{ m/s}$
9. QUESTION: An object is projected from ground level at 15.0° to the horizontal. Find the initial speed needed for a horizontal range of 2.50 m.
ANSWER: 7.00 m/s
SOLUTION:
$u^2 = \dfrac{Rg}{\sin 2\theta}$
$2\theta = 30.0^\circ$
$\sin 30.0^\circ = 0.500$
$u^2 = \dfrac{2.50 \times 9.8}{0.500}$
$u^2 = \dfrac{25\times10^{-1} \times 98\times10^{-1}}{5\times10^{-1}}$
$u^2 = 5 \times 98 \times10^{-1}$
$u^2 = 49.0\text{ m}^2/\text{s}^2$
$u = \sqrt{49.0}$
$u = 7.00\text{ m/s}$