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

D

Explanation:

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Answer 2

Explanation:

First, we'll find the height of the basketball at 1.7 seconds by plugging in t = 1.7 into the equation for the height of the ball:

h = 6 + 30t - 16t^2

h = 6 + 30(1.7) - 16(1.7)^2

h ≈ 24.32 feet

So the height of the basketball at 1.7 seconds is approximately 24.32 feet.

Next, we'll find the height of the leaping player's outstretched hands at 1.7 seconds by plugging in t = 1.7 into the equation for the height of the player's hands:

h = 9 + 25t - 16t^2

h = 9 + 25(1.7) - 16(1.7)^2

h ≈ 22.37 feet

So the height of the leaping player's outstretched hands at 1.7 seconds is approximately 22.37 feet.

Since the height of the basketball at 1.7 seconds (24.32 feet) is greater than the height of the leaping player's outstretched hands at 1.7 seconds (22.37 feet), the leaping player does not block the shot.