Question

modeling projectile motion
the projectile motion of an object can be modeled using ( h(t) = \frac{1}{2}gt^2 + v_0t + 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 ( -9.8 ) m/s².
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: Identify the formula and values

The height function is \( h(t)=\frac{1}{2}gt^{2}+v_{0}t + h_{0} \). We know \( g=- 9.8\) m/s², \( v_{0} = 20\) m/s, \( h_{0}=60\) m. When the object hits the ground, \( h(t) = 0\). So we substitute into the formula: \(0=\frac{1}{2}(-9.8)t^{2}+20t + 60\). Multiply both sides by 2 to eliminate the fraction: \(0=-9.8t^{2}+40t + 120\) (or we can work with the quadratic formula directly from \(h(t) = 0\) which is \( \frac{1}{2}gt^{2}+v_{0}t+h_{0}=0\), so \(gt^{2}+2v_{0}t + 2h_{0}=0\) when multiplied by 2. Wait, actually, the quadratic formula for \(ax^{2}+bx + c = 0\) is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Let's write \(h(t) = 0\) as \( \frac{1}{2}gt^{2}+v_{0}t+h_{0}=0\). Here, \(a=\frac{1}{2}g\), \(b = v_{0}\), \(c=h_{0}\). Given \(g=-9.8\), so \(a=\frac{1}{2}(-9.8)=- 4.9\), \(b = 20\), \(c = 60\). So the quadratic equation is \(-4.9t^{2}+20t + 60=0\). Now, using quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), substitute \(a=-4.9\), \(b = 20\), \(c = 60\). So \(t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(60)}}{2(-4.9)}\).

Step2: Match with the options

Let's check each option:

• First option: \(a = 4.9\) (wrong, since \(a=-4.9\))
• Second option: \(b = 60\) (wrong, \(b = 20\))
• Third option: \(a=-4.9\), \(b = 20\), \(c = 60\), so \(t=\frac{-20\pm\sqrt{(20)^{2}-4(-4.9)(60)}}{2(-4.9)}\) (matches)
• Fourth option: \(c=-60\) (wrong, \(c = 60\))

Answer:

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