A basketball player is shooting a basketball toward the net. The height, in feet, of the ball [tex]$t$[/tex] seconds after the shot is modeled by the equation [tex]$h=6+30 t-16 t^2$[/tex]. 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 [tex]$t$[/tex] seconds after he leaps is modeled by the equation [tex]$h=9+25 t-16 t^2$[/tex]. 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?
A. yes, exactly 0.6 seconds after the shot is launched
B. yes, between 0.64 seconds and 0.65 seconds after the shot is launched
C. yes, between 0.84 seconds and 0.85 seconds after the shot is launched
D. no, shot not blocked
Answer (2)
Set up the equations for the height of the ball and the blocker's hands as functions of time.
Equate the two height functions to find the time when the blocker's hands might intersect the ball's path.
Solve the resulting equation to find the time t ≈ 1.89 seconds.
Since 1.7"> t > 1.7 seconds, the shot is not blocked because the ball would have already reached the net. no, shot not blocked
Explanation
Understanding the Problem We are given the height of the basketball as a function of time: h b ( t ) = 6 + 30 t − 16 t 2 . The opposing player jumps to block the shot 0.2 seconds after the shot is launched. The height of the blocker's hands is given by h b l ( t ′ ) = 9 + 25 t ′ − 16 t ′2 , where t ′ is the time after the blocker jumps. We want to determine if the blocker blocks the shot, given that the ball is on a path to reach the net 1.7 seconds after the shot is launched.
Setting up the Equations Let t be the time after the shot is launched. The blocker jumps at t = 0.2 seconds. So, the time after the blocker jumps is t ′ = t − 0.2 . The height of the blocker's hands at time t is h b l ( t − 0.2 ) = 9 + 25 ( t − 0.2 ) − 16 ( t − 0.2 ) 2 for 0.2"> t > 0.2 .
Equating the Heights To determine if the shot is blocked, we need to find the time t when the height of the ball and the height of the blocker's hands are equal, i.e., h b ( t ) = h b l ( t − 0.2 ) . So we set up the equation:
6 + 30 t − 16 t 2 = 9 + 25 ( t − 0.2 ) − 16 ( t − 0.2 ) 2
Solving for Time Now, let's solve the equation for t :
6 + 30 t − 16 t 2 = 9 + 25 ( t − 0.2 ) − 16 ( t − 0.2 ) 2 6 + 30 t − 16 t 2 = 9 + 25 t − 5 − 16 ( t 2 − 0.4 t + 0.04 ) 6 + 30 t − 16 t 2 = 4 + 25 t − 16 t 2 + 6.4 t − 0.64 6 + 30 t = 3.36 + 31.4 t 2.64 = 1.4 t t = 1.4 2.64 = 1.8857
Determining if the Shot is Blocked The time t when the heights are equal is approximately 1.89 seconds. However, the ball is supposed to reach the net at 1.7 seconds. Since the time of potential block (1.89 seconds) is greater than the time when the ball reaches the net (1.7 seconds), the shot is not blocked.
Final Answer Therefore, the shot is not blocked.
Examples
Understanding projectile motion is crucial in sports like basketball. By modeling the height of a basketball shot with a quadratic equation, coaches and players can analyze the trajectory and optimize shooting techniques. For example, knowing the initial velocity and launch angle allows predicting whether a shot will be blocked or reach the basket, helping players adjust their shots for better accuracy. This blend of physics and mathematics enhances performance and strategic decision-making in the game.
The leaping player does not block the shot because the height of the blocker's hands reaches the same height as the basketball approximately 1.89 seconds after launch, while the ball is expected to reach the net at 1.7 seconds. Therefore, the shot is not blocked. The final answer is: D. no, shot not blocked.
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