Question

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tallplatform. The equation for the object's height s at time t seconds after launch is s(t) =-4.9t^2 +19.6t+ 58.8, where s is in meters.a) When does the object strike the ground?b) What is the initial height of the object?c) What is the maximum height of the object?

Answer 1

Solution

Step 1

$s(t)\text{ = -4.9t}^2+19.6t+58.8$

a)

The object strike the ground at s(t) = 0

$\begin{gathered} s(t)=\text{-4.9t}^2\text{+19.6t+58.8} \\ \\ -4.9t^2\text{{}+19.6t+58.8 = 0} \end{gathered}$$\begin{gathered} \mathrm{Multiply\:both\:sides\:by\:}10 \\ -4.9t^2\cdot \:10+19.6t\cdot \:10+58.8\cdot \:10=0\cdot \:10 \\ -49t^2+196t+588=0 \\ t_{1,\:2}=\frac{-196\pm \sqrt{196^2-4\left(-49\right)\cdot \:588}}{2\left(-49\right)} \\ t_{1,\:2}=\frac{-196\pm \:392}{2\left(-49\right)} \\ Separate\:the\:solutions \\ t_1=\frac{-196+392}{2\left(-49\right)},\:t_2=\frac{-196-392}{2\left(-49\right)} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=-2,\:t=6 \\ t\text{ cannot be negative} \\ t\text{ = 6 seconds} \end{gathered}$

b)

The initial height = 58.8

c)

$\begin{gathered} At\text{ maximum height, }\frac{ds(t)}{dt}=0 \\ \\ s(t)\text{ =}-4.9t^2+19.6t+58.8 \\ \\ \frac{ds(t)}{dt}=\text{ -9.8t + 19.6} \\ \\ -9.8t\text{ + 19.6 = 0} \\ \\ 9.8t\text{ = 19.6 } \\ \\ t\text{ =}\frac{19.6}{9.8}=2\text{ seconds} \\ \\ Maximum\text{ height =}-4.9\times2^2+19.6\times2+58.8 \\ \\ =-2^2\times \:4.9+39.2+58.8 \\ =78.4 \\ Maximum\text{ height = 78.4} \end{gathered}$