Question

question 5
the equation ( y = - 15 t ^ { 2 } + 60 t + 5 ) represents the height of a projectile, ( y ), in feet at a particular time ( t ), in seco
for what interval (or intervals) of time will the projectiles height be less than 19 feet?
between 1 and 4 seconds
between 0.25 and 3.5 seconds
less than 1 second and more than 4 seconds
less than 0.25 seconds and after 3.5 seconds
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concepts covered in this question

solving quadratic inequalities representing motion

note: these lide wont be here when you take the real milestone.

Explanation:

Step1: Set up the inequality

We want to find when \(y < 19\). Substitute \(y=-15t^{2}+60t + 5\) into the inequality:
\(-15t^{2}+60t + 5<19\)
Subtract 19 from both sides:
\(-15t^{2}+60t + 5-19<0\)
\(-15t^{2}+60t - 14<0\). Multiply through by \(- 1\) (and reverse the inequality sign):
\(15t^{2}-60t + 14>0\)

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 15\), \(b=-60\), \(c = 14\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-60)^{2}-4\times15\times14=3600 - 840=2760\)
\(t=\frac{60\pm\sqrt{2760}}{30}=\frac{60\pm2\sqrt{690}}{30}=\frac{30\pm\sqrt{690}}{15}\approx\frac{30\pm26.27}{15}\)
\(t_1=\frac{30 + 26.27}{15}\approx3.75\) (approximate value of \(\frac{30+\sqrt{690}}{15}\)), \(t_2=\frac{30 - 26.27}{15}\approx0.25\) (approximate value of \(\frac{30-\sqrt{690}}{15}\))

Since the parabola \(y = 15t^{2}-60t + 14\) (opens upwards, \(a=15>0\)), the solution of \(15t^{2}-60t + 14>0\) is \(t<0.25\) or \(t>3.5\)

Answer:

Less than \(0.25\) seconds and after \(3.5\) seconds