Fall

A body is thrown vertically upward at initial speed v₀ = 39 m/s. Its height as a function of time is given by the equation $h=v_0\cdot t-\frac{1}{2}\cdot 9.8\cdot t^2$. What is the maximum height reached?

Final Answer:

h =  77.6 m

Step-by-step explanation:

$$\begin{gathered} v_0=39\text{m/s} \\ g=9.8{\text{m/s}}^2 \\ {v}_0=g\cdot t \\ t=v_0/g=39/9.8=\frac{195}{49}\doteq 3.9796\text{s} \\ h=v_0\cdot t-\frac{1}{2}\cdot g\cdot t^2=39\cdot 3.9796-\frac{1}{2}\cdot 9.8\cdot 3.9796^2=\frac{7605}{98}\doteq 77.602\text{m} \\ \text{ Verifying Solution: } \\ h(t)=v_0\cdot t-\frac{1}{2}\cdot g\cdot t^2 \\ {h}^{\prime }(t)=\frac{dh(t)}{dt}=v_0-gt \\ {h}^{\prime }(t)=0 \\ {v}_0-gt=0 \\ {v}_0=gt \\ t=v_0/g \end{gathered}$$

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