Chapter P.7, Problem 33E

Solution

a.

The time when the projectiles height above ground be 768 ft.

After 4 seconds when going up and then after 12 second when it is coming down.

Given:

A projective is launched straight up from the ground level with an initial velocity of 256 ft/sec.

Concept Used:

• The height of an object lunched straight up with and initial velocity of v ft/sec is given by $h=-16t^2+vt$ .
• If there is the strict inequality symbol $<\text{ or}>$ , then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$ , then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.

Calculation:

Recall that the height of an object lunched straight up with and initial velocity of v ft/sec is given by $h=-16t^2+vt$ .

Since the projectile was launched at an initial velocity of 256 ft/sec, the height of the projectile after t seconds is given by $h=-16t^2+256t$ .

In order to find the time when the projectile height above the ground be 768 ft, solve the equation

  $\begin{array}{c}768=-16t^2+256t\\ 16t^2-256t+768=0\\ 16\left(t^2-16t+48\right)=0\\ t^2-16t+48=0\\ \left(t-4\right)\left(t-12\right)=0\\ \Rightarrow t=4,\text{ and }t=12\end{array}$

Thus, the height of the projectile will be 768 ft after 4 seconds when going up and then after 12 second when it is coming down.

b.

The time when the projectiles height above ground be at least 768 ft.

When t is in the interval $\left[4,12\right]$ .

Given:

A projective is launched straight up from the ground level with an initial velocity of 256 ft/sec.

Concept Used:

• The height of an object lunched straight up with and initial velocity of v ft/sec is given by $h=-16t^2+vt$ .
• If there is the strict inequality symbol $<\text{ or}>$ , then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$ , then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.

Calculation:

Since the height of the projectile is 768 ft above ground after 4 seconds of launch and till 12 second when it is coming down.

Hence, the projectile’s height above the ground be at least 768 ft for the interval $\left[4,12\right]$ .

c.

The time when the projectiles height above ground be less than or equal to 768 ft.

When t is in the interval $\left(0,4\right]\cup \left[12,16\right)$ .

Given:

A projective is launched straight up from the ground level with an initial velocity of 256 ft/sec.

Concept Used:

• The height of an object lunched straight up with and initial velocity of v ft/sec is given by $h=-16t^2+vt$ .
• If there is the strict inequality symbol $<\text{ or}>$ , then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$ , then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.

Calculation:

Since the height of the projectile is 768 ft above ground after 4 seconds of launch and till 12 second when it is coming down.

Note that the projectile will hit the ground when height is 0, that is

  $\begin{array}{c}0=-16t^2+256t\\ 16t^2-256t=0\\ 16\left(t^2-16t\right)=0\\ t^2-16t=0\\ t\left(t-16\right)=0\\ \Rightarrow t=0,\text{ and }t=16\end{array}$

Hence, the projectile’s height above the ground be less than or equal to 768 ft for the interval $\left(0,4\right]\cup \left[12,16\right)$ .