479,282 views
17 votes
17 votes
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 ?

User Robertoplancarte
by
2.9k points

1 Answer

26 votes
26 votes

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=(-b\pm√(b^2-4ac))/(2a)

Plug the values


\begin{gathered} t=(-25\pm√(25^2-4\cdot(-1)\cdot150))/(2\cdot(-1)) \\ \\ t=(-25\pm√(625+4\cdot150))/(-2) \\ \\ t=(25\pm√(625+600))/(2) \\ \\ t=(25\pm√(1225))/(2) \\ \\ t=(25\pm35)/(2) \\ \\ t=(25+35)/(2)=(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

User Jap Mul
by
2.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.