Science:Math Exam Resources/Courses/MATH110/December 2016/Question 06 (b)/Solution 1

Using the constant $g={\frac {3}{4}}$ found in part (a), the height of the ball is given by

$$s(t)=15t-{\frac {3}{8}}t^{2}$$

As suggested in the hint, the velocity is a derivative of the height function;

$$v(t)=s'(t)=\left(15t-{\frac {3}{8}}t^{2}\right)'=15-{\frac {3}{4}}t$$

.

By plugging $t=0$, we get a $y$-intercept as $y=15$. On the other hand, solving $v(t)=15-{\frac {3}{4}}t=0$, we have a $t$-intercept $t=20$.

Since its graph is a straight line with slope $-{\frac {3}{4}}$ passing through the intercepts $(0,15)$ and $(20,0)$, we can sketch it as follows.