Question

justin and elena each launched a toy rocket into the air. the height of justins rocket is modeled by the equatic $16t^{2}+60t + 2$. elena launched his rocket from the same position, but with an initial velocity double that of justin which equation best models the height of elenas rocket? $h(t)=at^{2}+vt + h_{0}$ $h=-16t^{2}+60t + 4$ $h=-32t^{2}+120t + 4$ $h=-32t^{2}+60t + 2$ $h=-16t^{2}+120t + 2$

Explanation:

Step1: Identify the initial velocity and launch position of Justin's rocket

The height formula is \(h(t)=at^{2}+vt + h_{0}\). For Justin's rocket \(h(t)=- 16t^{2}+60t + 2\), so \(a=-16\) (acceleration due to gravity, constant for both rockets), \(v = 60\) (initial velocity of Justin's rocket), \(h_{0}=2\) (launch - height, same for both rockets).

Step2: Calculate the initial velocity of Elena's rocket

Elena's initial velocity is double that of Justin's. So \(v_{Elena}=2\times60 = 120\). The launch - height \(h_{0}\) remains the same (\(h_{0}=2\)) and the acceleration \(a=-16\) (since the acceleration due to gravity is constant near the Earth's surface for both rockets).

Answer:

\(h=-16t^{2}+120t + 2\) (the fourth option)