Chapter 1.8, Problem 74PS

Solution

a)

To find: $v_0$ using the fact that the maximum value of $h$ is $921+160=1081$ feet

Value of $v_0$ is $6.325$ feet per second.

Given:

The Stratosphere Tower in Las Vegas is $921$ feet tall and has a "needle" at its top that extends even higher into the air. A thrill ride called the Big Shot catapults riders $160$ feet up the needle and then lets them fall back to the launching pad. The height $h$ (in feet) of a rider on the Big Shot can be modeled by $h=-16t^2+v_{0}t+921$ where $t$ is the elapsed time (in seconds) after launch and $v_0$ is the initial vertical velocity (in feet per second).

Calculation:

The height $h$ (in feet) of a rider on the Big Shot can be modeled by $h=-16t^2+v_{0}t+921$

where $t$ is the elapsed time (in seconds) after launch and $v_0$ is the initial vertical velocity (in feet per second).

Find $v_0$ using the fact that the maximum value of $h$ is $921+160=1081$ feet .

Rewrite the model of the height $h=-16t^2+v_{0}t+921$ as a vertex form so that the fact that the maximum value of $h$ is $921+160=1081$ feet can be used.

  $\begin{array}{c}h=-16t^2+v_{0}t+921\\ =-16\left(t^2-v_{0}t+10\right)+1081\\ =-16\left(t^2-2\sqrt{10}t+10\right)+1081\\ =-16{\left(t-\sqrt{10}\right)}^2+1081\end{array}$

The vertex form $h=-16{\left(t-\sqrt{10}\right)}^2+1081$ for the height shows that the maximum height is $921+160=1081$ feet.

Hence, the initial vertical velocity, $v_0$ , is $2\sqrt{10}$ feet per second or about $6.325$ feet per second

b)

To compare: Time taken by needle with the time given by the model $h=-16t^2+v_{0}t+921$ where $v_0$ is the value you found in part $\left(a\right)$

The vertex form $h=-16{\left(t-\frac{1}{16}\right)}^2+921.0625$ for the height shows that the maximum height of $h$ is $921.0625$ feet which is not same as we did in the part $\left(a\right)$ .

Given:

The Stratosphere Tower in Las Vegas is $921$ feet tall and has a "needle" at its top that extends even higher into the air. A thrill ride called the Big Shot catapults riders $160$ feet up the needle and then lets them fall back to the launching pad. A brochure for the Big Shot states that the ride up the needle takes two seconds.

Calculation:

Substitute $2$ seconds for $v_0$ in the model $h=-16t^2+v_{0}t+921$

  $\begin{array}{c}h=-16t^2+v_{0}t+921\\ =-16t^2+2t+921\\ =-16\left[t^2-\frac{1}{8}t+{\left(\frac{1}{16}\right)}^2\right]+\frac{14737}{16}\\ \approx -16\left[t^2-\frac{1}{8}t+{\left(\frac{1}{16}\right)}^2\right]+921.0625\end{array}$

So,

  $h=-16{\left(t-\frac{1}{16}\right)}^2+921.0625$

The vertex form $h=-16{\left(t-\frac{1}{16}\right)}^2+921.0625$ for the height shows that the maximum height of h is $921.0625$ feet which is not same as we did in the part $\left(a\right)$ .