Science:Math Exam Resources/Courses/MATH110/April 2018/Question 05 (c)

MATH110 April 2018

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Question 05 (c)
A stone is shot vertically into the air at an initial velocity of 16 m/s. Its height $s$ (in metres) above the ground after $t$ seconds is given by $$s(t)=16t-5t^{2}.$$ (c) Suppose the same stone is shot vertically on Mars. On Mars the height of the stone above ground after $t$ seconds would be $$s(t)=16t-2t^{2}.$$ How much higher will the stone travel on Mars than on Earth?

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Hint
Find the maximum value of the stone heights on Mars and on Earth. For the maximum value on Earth, we can use the part (a).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By part (a), the maximum of the stone height on Earth occurs at $t={\frac {8}{5}}$, so the maximum value is $$s\left({\frac {8}{5}}\right)=16\cdot {\frac {8}{5}}-5\left({\frac {8}{5}}\right)^{2}={\frac {8}{5}}\left(16-5\cdot {\frac {8}{5}}\right)={\frac {8}{5}}\cdot 8={\frac {64}{5}}.$$ Now, we find the maximum value of the stone height on Mars. As we did in part (a), the maximum occurs at $t_{1}>0$ satisfying $s'(t_{1})=0$. Since the derivative of the stone height on Mars is $$s'(t)=(16t-2t^{2})'=16-4t=4(4-t),$$ the value $t_{1}$ is $4$. Therefore, the maximum on Mars is $$s(4)=16\cdot 4-2\cdot 4^{2}=4(16-8)=32.$$ Finally we get the difference between two maximums as $${\text{(Maximum on Mars)}}-{\text{(Maximum on Earth)}}=32-{\frac {64}{5}}={\frac {160-64}{5}}={\frac {96}{5}}.$$ Answer: $\color {blue}{\frac {96}{5}}$ (metres)