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?
Answers
Answer:
Blocker can not block the ball.
Step-by-step explanation:
Shooter shoots the ball and height function is
h = 6 + 30t – 16*t*t
This is applicable from t = 0.
Blocker jumps at t = 2/10 sec and the height function is
h = 9 + 25t - 16*t*t
Blocker starts 0.2 seconds after the ball was shot.
Ball reaches net in 1.7 seconds.
Height of the ball in 1.7 seconds is defined by first equation.
h = 6 + 30t – 16*t*t = 6 + 30*1.7 – 16*1.7*17 = 10.76 Units.
Blocker jumps and his height when the ball reaches net is defined by second equation.
But instead of t = 1.7 seconds, it is delayed by 0.2 seconds. i.e. t = 1.5 seconds.
h = 9 + 25t - 16*t*t = 9 + 25*1.5 – 16*1.5*1.5 = 10.5 units.
The blocker can only jump 10.5 unit height where as the ball is at 10.76units height at 1.7 seconds.
Hence blocker can not block the ball.
Answer:
D. not, shot was blocked. Edg 2021
Step-by-step explanation:
took test :)