Math  /  Calculus

Question16. The acceleration due to gravity on Earth is 9.8 m/s29.8 \mathrm{~m} / \mathrm{s}^{2}. A ball is thrown upward at an initial velocity of 15 m/s15 \mathrm{~m} / \mathrm{s} from a height of 1 m above the ground. Round answers to the nearest tenth. a) Write an equation for the height of the ball. b) What is the height of the ball after 1 s ? c) After how many seconds does the ball land? d) What is the maximum height of the 20. ball? When does this occur? e) Repeat parts a) to d) for a ball thrown on the Moon, where g=1.62 m/s2g=1.62 \mathrm{~m} / \mathrm{s}^{2}. f) Repeat parts a) to d) for a ball thrown on Jupiter, where g=23.1 m/s2g=23.1 \mathrm{~m} / \mathrm{s}^{2}.

Studdy Solution
Calculate the height after 1 second on Jupiter: h(1)=1+15(1)11.55(1)2 h(1) = 1 + 15(1) - 11.55(1)^2 h(1)=4.45 h(1) = 4.45
Time when the ball lands on Jupiter: 0=1+15t11.55t2 0 = 1 + 15t - 11.55t^2 11.55t215t1=0 11.55t^2 - 15t - 1 = 0
Use the quadratic formula: t=15±225+46.223.1 t = \frac{15 \pm \sqrt{225 + 46.2}}{23.1} t15+16.4823.1 t \approx \frac{15 + 16.48}{23.1} t1.4 t \approx 1.4
Maximum height on Jupiter: t=15211.55 t = \frac{-15}{2 \cdot 11.55} t0.65 t \approx 0.65
h(0.65)=1+15(0.65)11.55(0.65)2 h(0.65) = 1 + 15(0.65) - 11.55(0.65)^2 h(0.65)5.9 h(0.65) \approx 5.9
The solutions for each scenario are: - Earth: Height after 1 s: 11.1 m, Lands after 3.1 s, Max height: 12.5 m at 1.53 s. - Moon: Height after 1 s: 15.19 m, Lands after 18.6 s, Max height: 70.4 m at 9.26 s. - Jupiter: Height after 1 s: 4.45 m, Lands after 1.4 s, Max height: 5.9 m at 0.65 s.

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