Question

modeling projectile motion
the projectile motion of an object can be modeled using $h(t)=\frac{1}{2}gt^{2}+v_{0}t + h_{0}$, where $g$ is the acceleration due to gravity, $t$ is the time since launch, $h(t)$ is the height of the projectile after time $t$, $v_{0}$ is the initial velocity, and $h_{0}$ is the initial height. the acceleration due to gravity is $-9.8m/s^{2}$. an object is launched at an initial velocity of 20 meters per second and an initial height of 60 meters.
which solution can be simplified to determine when the object will hit the ground?
$t=\frac{-20\pm\sqrt{(20)^{2}-4(4.9)(60)}}{2(4.9)}$
$t=\frac{-60\pm\sqrt{(60)^{2}-4(-4.9)(20)}}{2(-4.9)}$
$t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(60)}}{2(-4.9)}$
$t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(-60)}}{2(-4.9)}$

Explanation:

Step1: Substitute values into the height formula

When the object hits the ground, \(h(t) = 0\). Given \(g=-9.8\), \(v_0 = 20\), \(h_0=60\), the equation \(h(t)=\frac{1}{2}gt^{2}+v_0t + h_0\) becomes \(0=\frac{1}{2}(-9.8)t^{2}+20t + 60\), which simplifies to \(-4.9t^{2}+20t + 60=0\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here \(a=-4.9\), \(b = 20\), \(c = 60\). Substituting these values into the quadratic formula gives \(t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(60)}}{2(-4.9)}\).

Answer:

\(t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(60)}}{2(-4.9)}\) (the third option)