Question

justin and elena each launched a toy rocket into the air. the height of justin’s rocket is modeled by the equation $h = -16t^2 + 60t + 2$. elena launched his rocket from the same position, but with an initial velocity double that of justin’s. which equation best models the height of elena’s rocket?
$h(t) = at^2 + vt + h_0$
$\bigcirc\\ h = -16t^2 + 60t + 4$
$\bigcirc\\ h = -32t^2 + 120t + 4$
$\bigcirc\\ h = -32t^2 + 60t + 2$
$\bigcirc\\ h = -16t^2 + 120t + 2$

Explanation:

Step1: Identify parameters in Justin's equation

Justin's height equation: \( h = -16t^2 + 60t + 2 \). Comparing with \( h(t)=at^2 + vt + h_0 \), we have \( a=-16 \), \( v = 60 \) (initial velocity), \( h_0 = 2 \) (initial height).

Step2: Determine Elena's parameters

Elena launches from the same position, so \( h_0 \) remains \( 2 \). Her initial velocity is double Justin's, so new \( v = 2\times60 = 120 \). The acceleration \( a \) (due to gravity) remains the same? Wait, no—wait, in the standard projectile motion, the quadratic coefficient \( a \) is related to gravity. Wait, but in Justin's equation, \( a=-16 \). Wait, maybe the problem is considering the model \( h(t)=at^2 + vt + h_0 \), where \( a \) is the acceleration (here, -16, maybe ft/s²), \( v \) is initial velocity, \( h_0 \) initial height. Since Elena launches from the same position, \( h_0 = 2 \). Initial velocity \( v \) is doubled: \( v = 2\times60 = 120 \). The \( a \) term: wait, is \( a \) changing? Wait, no—wait, maybe the problem has a typo, but looking at the options, let's re - evaluate. Wait, Justin's equation: \( h=-16t^2 + 60t + 2 \). Elena's initial velocity is double, so \( v \) becomes \( 120 \), initial height \( h_0 \) is same (2), and what about \( a \)? Wait, maybe the problem is considering that the quadratic coefficient (acceleration) is also doubled? No, that doesn't make sense. Wait, no—wait, looking at the options, let's check the options. Wait, the correct approach: same initial position (\( h_0 = 2 \)), double initial velocity (\( v = 120 \)), and the \( a \) term: in Justin's equation, \( a=-16 \). Wait, but maybe the problem is that the acceleration is also doubled? No, that's not physical. Wait, no—wait, the options: let's see. The model is \( h(t)=at^2 + vt + h_0 \). Justin: \( a=-16 \), \( v = 60 \), \( h_0 = 2 \). Elena: same \( h_0 = 2 \), \( v = 2\times60 = 120 \), and what about \( a \)? Wait, maybe the problem has a mistake, but looking at the options, the correct one should have \( h_0 = 2 \), \( v = 120 \), and \( a=-16 \)? Wait, no, one of the options is \( h=-16t^2 + 120t + 2 \). Let's check:

Justin's equation: \( h=-16t^2 + 60t + 2 \). Elena: same initial position (\( h_0 = 2 \)), double initial velocity (\( v = 120 \)), and the acceleration term \( a \) remains - 16 (since gravity doesn't change). So Elena's equation is \( h=-16t^2+120t + 2 \).

Answer:

\( h=-16t^2 + 120t + 2 \) (the last option: \( h=-16t^2 + 120t + 2 \))