Chapter ISG, Problem 8.6.2.4P

a.

To determine

To write an equation to express the time that the stuntman was in the air.

Answer to Problem 8.6.2.4P

The equation to express the time that the stuntman was in air is $-16t^2+16t+3=0$ .

Explanation of Solution

Given information :

Given function is $h=-16t^2+vt+h_0$ .

  $v=16ft/s$

  $h_0=3$ feet.

The determine the time that the stuntman was in the air, replace the velocity variable in the function given equation, along with the initial height.

This would yield an equation that would be showing the relationship of time along with the height covered.

Here, replace $h_0$ and $v$ with their numeric values.

  $v=16ft/s$

  $h_0=3$ feet.

Hence, the resulting function is $f\left(t\right)=-16t^2+16t+3$ . Equation is $-16t^2+16t+3=0$ .

b.

To determine

To calculate the number of seconds that the stuntman was in the air before touching the ground.

Answer to Problem 8.6.2.4P

The stuntman was in the air for 1.16 seconds .

Explanation of Solution

Given information :

Given function is $h=-16t^2+vt+h_0$ .

Formula used :

Put the variable in the given function and derive a quadratic equation. Then, solve for the roots that is the time t.

Use formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to solve the same.

Calculation :

Given function: $h=-16t^2+vt+h_0$

  $v=16ft/s$

  $h_0=3$

Thus, equation for t is $-16t^2+16t+3=0$

Here, $a=-16$ , $b=16$ and $c=3$ .

  $t=\frac{-16\pm \sqrt{{16}^2-4\times -16\times 3}}{-2\times 16}$

Putting values in the formula used.

  $=\frac{-16\pm \sqrt{256+192}}{-32}$

Simplifying the same.

  $=\frac{-16\pm \sqrt{448}}{-32}$

This can be further simplified as.

  $=\frac{-16\pm 21.16}{-32}$

Applying square root.

  $=\frac{-16-21.16}{-32}$

Solving one of the root.

  $=\frac{37.16}{32}$

This will give

  $=1.16$ seconds.

  $=\frac{-16+21.16}{-32}$

Solving the other root.

  $=-\frac{5.16}{32}$

This will give

  $=-0.161$ seconds.

Since the time cannot be negative, the time taken for by the stuntman to reach the ground is 1.16 seconds .

c.

To determine

To calculate the change of calculation if the stuntman was launched from a height of more than 3 feet.

Answer to Problem 8.6.2.4P

The equation to would change the change of the constant $h_0$ , so function will be $f\left(t\right)=-16t^2+16t+h_0$ .

Explanation of Solution

Given information :

Given function is $h=-16t^2+vt+h_0$ .

  $v=16ft/s$

The determine the time that the stuntman was in the air, replace the velocity variable in the function given equation, along with the initial height.

This would yield an equation that would be showing the relationship of time along with the height covered.

Here, replace $h_0$ and $v$ with their numeric values.

  $v=16ft/s$

Since $h_0<3$ , change this value to the new height and calculate the time hence taken, as shown in sub-part b.

Thus the function would be $h=-16t^2+vt+h_0$ .