Question

A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modeled by the equation h = 6 + 30t – 16t2. Two-tenths of a second after the shot is launched, an opposing player leaps up to block the shot. The height of the shot blocker’s outstretched hands t seconds after he leaps is modeled by the equation h = 9 + 25t – 16t2. If the ball reaches the net 1.7 seconds after the shooter launches it, does the leaping player block the shot?
Yes, exactly 0.6 seconds after the shot is launched.
Yes, between 0.64 seconds and 0.65 seconds after the shot is launched
Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.
No, the shot is not blocked.

Answer 1

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Step-by-step explanation:

$\text{The equation for ball height} : 6+30\cdot t-16\cdot t^{2}\\\text{The equation for shot blocker's height} : 9+25\cdot t-16\cdot t^{2}$

But, the shot is made before two tenths of a second or 0.2 seconds therefore modified equation for ball height is :

$6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}$

Now for the shot to be blocked,the height of shot blocker must be greater than the height of the ball which is shot before 0.2 seconds :

$\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}\\\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot t-6-16\cdot (t^{2}-0.4\cdot t+0.04)\\\implies9+25\cdot t-16\cdot t^{2}\geq 30\cdot t-16\cdot t^{2}+6.4\cdot t-0.64\\\implies 9+25\cdot t\geq 36.4\cdot t-0.64\\\implies 9.64\geq 11.4\cdot t\\\\\implies t\leq \frac{9.64}{11.4}\approx 0.846$

Hence, the shot was blocked between 0.84 and 0.85 seconds after the shot is launched.

Answer 2

Answer:

The correct answer on edgen is D) NO, SHOT NOT BLOCKED

Step-by-step explanation:

Keeping It Simple.