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 was on a path to reach 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, shot not blocked.

Answer 1

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

what will be the time in which the blocker will block the ball?

The given equation for ball height

$=6+30t-16t^{2}$

The equation for the blocker's height will be

$=9+25t-16t^{2}$

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

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

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

$9+25t-16t^{2} \geq 6+30(t-0.2)-16(t-0.2)^{2}$

$9+25t-16t^{2} \geq6+30t-6-16(t^{2} -0.4t+0.04)$

$9+25t-16t^{2} \geq30t-16t^{2} -6.4t+0.64$

$9+25t\geq36.4t-0.64$

$9.64\geq11.4t$

$t \leq \dfrac{9.64}{11.4}=0.846$

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

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