Chapter P.7, Problem 34E

Solution

a.

The time when the projectiles height above ground be 960 ft.

After 5 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 272 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 272 ft/sec, the height of the projectile after t seconds is given by $h=-16t^2+272t$ .

In order to find the time when the projectile height above the ground be 960 ft, solve the equation

  $\begin{array}{c}960=-16t^2+272t\\ 16t^2-272t+960=0\\ 16\left(t^2-17t+60\right)=0\\ t^2-17t+60=0\\ \left(t-5\right)\left(t-12\right)=0\\ \Rightarrow t=5,\text{ and }t=12\end{array}$

Thus, the height of the projectile will be 960 ft after 5 seconds when going up and then after 12 second when it is coming down.

b.

The time when the projectiles height above ground be more than 960 ft.

When t is in the interval $\left(5,12\right)$ .

Given:

A projective is launched straight up from the ground level with an initial velocity of 272 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 960 ft above ground after 5 seconds of launch and till 12 second when it is coming down.

Hence, the projectile’s height above the ground be more than 960 ft for the interval $\left(5,12\right)$ .

c.

The time when the projectiles height above ground be less than or equal to 960 ft.

When t is in the interval $\left(0,5\right]\cup \left[12,17\right)$ .

Given:

A projective is launched straight up from the ground level with an initial velocity of 272 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 960 ft above ground after 5 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

  $\begin{array}{c}0=-16t^2+272t\\ 16t^2-272t=0\\ 16\left(t^2-17t\right)=0\\ t^2-17t=0\\ t\left(t-17\right)=0\\ \Rightarrow t=0,\text{ and }t=17\end{array}$

Hence, the projectile’s height above the ground be less than or equal to 960 ft for the interval $\left(0,5\right]\cup \left[12,17\right)$ .