Question

Physicists tell us that altitude h in feet of a projectilet seconds after firing is h=-1672 +8/+he,where y, is the initial velocity in feet per second and h, is the altitude in feet from which it is fired. If arocket is launched from a hilltop 2400 feet above the desert with an initial upward velocity of 400 feet persecond, then when will it land on the desert?

Answer 1

The projectile will land on the desert when h = 0, in other words

`0=-16t^2+v_0t+h_0`

Now, we are told that h_o = 2400ft and v_o = 400ft/s; and putting those in the formula above gives

`-16t^2+400_{}t+2400=0`

Now this is a quadratic equation and the solution is given by

`\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

In our case, a = -16, b = 400, and c = 2400; therefore, the quadratic formula above gives

`t=\frac{-400\pm\sqrt[]{400^2-4(-16)(2400)}}{2(-16)}`
`t=(-400\pm560)/(-32)`

Hence, the two solutions are

`\begin{gathered} t=30 \\ t=-5 \end{gathered}`

Since negative values of t have no physical significance, the valid answer is t = 30s.

Hence, the third choice in the column is the right answer.