Question

An onject is thrown upward from a height of 640ft with a velocity of 96ft/s. How long for the object to reach the ground?

Answer 1

The function h is given by:

`\begin{gathered} h(t)=-16t^2+v_0t+h_0 \\ \text{ Substitute }v_0=96\text{ and }h_0=640\text{ into the equation:} \\ \end{gathered}`

Therefore,

`h(t)=-16t^2+96t+640`

When the object reaches the ground, h(t)=0:

`-16t^2+96t+640=0`

Divide both sides by -16:

`\begin{gathered} t^2-6t-40=0 \\ t^2+4t-10t-40=0 \\ t(t+4)-10(t+4)=0 \\ (t-10)(t+4)=0 \\ \text{ Therefore,} \\ t=10,-4 \end{gathered}`

Since t is not negative, discard the negative value of t.

Therefore, the required time is t = 10s