Question

Physicists tell us that altitude h in feet of a projectile t seconds after firing is h=-16t^2+v0t+h0, where v0 is the initial velocity in feet per second and h0 is the altitude in feet from which it is fired. If a rocket is launched from a hilltop 2400 feet above the desert with an initial upward velocity of 400 feet per second, then when will it land on the desert ?

Answer 1

The formula that the physicists told was

$h(t)=-16t^2+v_0t+h_0$

We know that

v₀ = 400 ft/s

h₀ = 2400 ft

Then let's put it into the formula

$\begin{gathered} h(t)=-16t^2+400t+2400 \\ \\ \end{gathered}$

We want to know then it will land on the desert, in other words, when the height is equal to zero, then

$-16t^2+400t+2400=0$

Then we must solve that quadratic equation, to solve it let's first divide all by 16

$\begin{gathered} -16t^2+400t+2400=0 \\ \\ -t^2+25t+150=0 \end{gathered}$

Because it's an easier equation to solve and the solution is the same. Now we can apply the quadratic formula

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plug the values

$\begin{gathered} t=\frac{-25\pm\sqrt{25^2-4\cdot(-1)\cdot150}}{2\cdot(-1)} \\ \\ t=\frac{-25\pm\sqrt{625+4\cdot150}}{-2} \\ \\ t=\frac{25\pm\sqrt{625+600}}{2} \\ \\ t=\frac{25\pm\sqrt{1225}}{2} \\ \\ t=\frac{25\pm35}{2} \\ \\ t=\frac{25+35}{2}=\frac{60}{2}=30\text{ seconds} \end{gathered}$

We can ignore the other solution because it's negative and negative time is not a valid solution. Therefore, 30 seconds after its launch, the rocket will land in the desert