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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.

Respuesta :

nobillionaireNobley
For the shot to be blocked
9 + 25t - 16t^2 ≥ 6 + 30(t - 0.2) - 16(t - 0.2)^2
9 + 25t - 16t^2 ≥ 6 + 30t - 6 - 16(t^2 - 0.4t + 0.04)
9 + 25t - 16t^2 ≥ 30t - 16t^2 + 6.4t - 0.64
9 + 25t ≥ 36.4t - 0.64
9 + 0.64 ≥ 36.4t - 25t
9.64 ≥ 11.4t
t ≤ 9.64/11.4
t ≤ 0.846

Therefore, The shot was blocked between 0.84 and 0.85 seconds after the shot is launched.
0003145

Answer:

D

Step-by-step explanation:

h_1=6+30(t+0.2)-16(t+0.2)^2

h_2=6+30t-16t^2

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