Question

a projectile is fired straight up from ground level with an initial velocity of 112 ft/s. its height, h, above the ground after t seconds is given by ( h = -16t^2 + 112t ). what is the interval of time during which the projectile’s height exceeds 192 feet?

• ( 3 < t < 4 )
• ( t < 4 )
• ( t > 4 )
• ( 3 < t > 4 ) (note: likely a typo, probably ( 3 < t < 4 ) or similar intended, but as per ocr, this is the option)

Explanation:

Step1: Set up the inequality

We need to find when \( h>192 \), so substitute \( h = - 16t^{2}+112t \) into the inequality:
\( -16t^{2}+112t>192 \)
Subtract 192 from both sides to get a quadratic inequality:
\( -16t^{2}+112t - 192>0 \)
Divide all terms by - 16 (remember to reverse the inequality sign):
\( t^{2}-7t + 12<0 \)

Step2: Factor the quadratic

Factor \( t^{2}-7t + 12 \):
We need two numbers that multiply to 12 and add to - 7. The numbers are - 3 and - 4.
So, \( t^{2}-7t + 12=(t - 3)(t - 4) \)
The inequality becomes \( (t - 3)(t - 4)<0 \)

Step3: Analyze the sign of the quadratic

To solve \( (t - 3)(t - 4)<0 \), we find the critical points by setting each factor equal to zero: \( t - 3 = 0\) gives \( t = 3 \), and \( t - 4=0 \) gives \( t = 4 \).
These critical points divide the number line into three intervals: \( (-\infty,3) \), \( (3,4) \), and \( (4,\infty) \).

• For \( t<3 \) (e.g., \( t = 0 \)): \( (0 - 3)(0 - 4)=(-3)\times(-4) = 12>0 \), so the inequality is not satisfied.
• For \( 3
• For \( t>4 \) (e.g., \( t = 5 \)): \( (5 - 3)(5 - 4)=(2)\times(1) = 2>0 \), so the inequality is not satisfied.

Answer:

\( 3 < t < 4 \) (assuming the first option is \( 3 < t < 4 \) and the last option has a typo, the correct interval is \( 3 < t < 4 \))