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

MATH110 April 2018

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Question 05 (b)
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 $$s(t)=16t-5t^{2}.$$ (b) What is the velocity of the stone when it hits the ground?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, find the time at which the stone hits the ground.

Hint 2
Note that the velocity $v$ is a derivative of the height function $s$. i.e., $v(t)=s'(t)$.

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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 Hint 1, we first solve $s(t)=0$. Since the factorization of $s$ is $s(t)=t(16-5t)=5t\left({\frac {16}{5}}-t\right)$, we can easily find that the solutions to $s(t)=0$ are $t=0$ and $t={\frac {16}{5}}.$ Since $t=0$ is the starting time, the stone hits the ground at $t={\frac {16}{5}}$. On the other hand, by Hint 2 and the solution of part (a), the velocity is $$v(t)=s'(t)=16-10t.$$ Therefore, the velocity at $t={\frac {16}{5}}$ is $$v\left({\frac {16}{5}}\right)=16-10\cdot {\frac {16}{5}}=16-32=-16.$$ Answer: $\color {blue}-16$ (metre per sec)