Question

1. calculator

a cannonball is shot from a cannon into the air with an upward velocity of 40 ft/sec. the equation that gives the height (h) in feet of the ball at any time (t) in seconds is: h(t)= -16t² + 40t + 1.5.
● what is the initial height of the cannon ball?
● what is the maximum height of the cannon ball?
● after how many seconds does the cannon ball reach its maximum height?
● how long does it take for the cannon ball to hit the ground?

Explanation:

What is the initial height of the cannon ball?

Step1: Substitute \(t = 0\) into the height function

The height function is \(h(t)=-16t^{2}+40t + 1.5\). When \(t = 0\), we have \(h(0)=-16\times0^{2}+40\times0 + 1.5\).

Step2: Simplify the expression

Since \(-16\times0^{2}=0\) and \(40\times0 = 0\), then \(h(0)=1.5\) feet.

What is the maximum height of the cannon ball?

Step1: Find the time \(t\) at which the maximum occurs

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 40\), \(c = 1.5\)), the time \(t\) at the vertex (maximum for \(a<0\)) is given by \(t=-\frac{b}{2a}\). Substitute \(a=-16\) and \(b = 40\) into the formula: \(t=-\frac{40}{2\times(-16)}=\frac{40}{32}=\frac{5}{4}=1.25\) seconds.

Step2: Substitute \(t = 1.25\) into the height function

\(h(1.25)=-16\times(1.25)^{2}+40\times1.25 + 1.5\). First, calculate \((1.25)^{2}=1.5625\), then \(-16\times1.5625=-25\), \(40\times1.25 = 50\). So \(h(1.25)=-25 + 50+1.5=26.5\) feet.

After how many seconds does the cannon ball reach its maximum height?

Step1: Use the vertex - time formula

For the quadratic function \(h(t)=-16t^{2}+40t + 1.5\) (where \(a=-16\), \(b = 40\)), the time \(t\) at the vertex is \(t=-\frac{b}{2a}\). Substitute \(a=-16\) and \(b = 40\) into the formula: \(t=-\frac{40}{2\times(-16)}=\frac{40}{32}=\frac{5}{4}=1.25\) seconds.

How long does it take for the cannon ball to hit the ground?

Step1: Set \(h(t)=0\)

We have the equation \(-16t^{2}+40t + 1.5=0\). Multiply through by \(- 2\) to get \(32t^{2}-80t - 3=0\).

Step2: Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Here \(a = 32\), \(b=-80\), \(c=-3\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-80)^{2}-4\times32\times(-3)=6400 + 384=6784\). Then \(\sqrt{\Delta}=\sqrt{6784}=82.37\) (approx).
\(t=\frac{80\pm82.37}{64}\). We have two solutions: \(t_{1}=\frac{80 + 82.37}{64}\approx2.54\) and \(t_{2}=\frac{80-82.37}{64}\approx - 0.04\). Since time cannot be negative, \(t\approx2.54\) seconds.

Answer:

• Initial height: \(1.5\) feet.
• Maximum height: \(26.5\) feet.
• Time to reach maximum height: \(1.25\) seconds.
• Time to hit the ground: Approximately \(2.54\) seconds.