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Answer :
(a) The acceleration of the basketball at the highest point of its trajectory is 9.8 m/[tex]s^2.[/tex]
(b) The shooting guard must shoot the basketball at a speed of approximately 9.64 m/s to go through the hoop without touching the backboard.
a) To determine the acceleration of the basketball at the highest point of its trajectory, we can use the fact that the acceleration due to gravity is constant and equal to 9.8 m/[tex]s^2[/tex].
Since the ball is at its highest point, it has zero vertical velocity.
Therefore, the acceleration of the basketball at the highest point is equal to the acceleration due to gravity, which is 9.8 m/[tex]s^2[/tex].
b) To determine the speed at which the player must shoot the basketball so that it goes through the hoop without touching the backboard, we can use the kinematic equation for projectile motion.
The horizontal distance from the shooting guard to the hoop is 9.5 m.
The vertical distance from the shooting guard's hand to the hoop is 3.05 m - 1.95 m = 1.1 m.
We can use the equation:
Range = (initial velocity)² * sin(2*theta) / acceleration due to gravity.
We know the range is 9.5 m, the vertical displacement is 1.1 m, the angle is 48.5°, and the acceleration due to gravity is 9.8 m/s^2.
Rearranging the equation to solve for initial velocity, we have:
(initial velocity)² = Range * acceleration due to gravity / sin(2*theta)
Plugging in the values, we get:
(initial velocity)² = 9.5 * 9.8 / sin(97°)
Simplifying the expression gives:
(initial velocity)² = 93.1
initial velocity = [tex]\sqrt{93.1}[/tex]
initial velocity = 9.64 m/s.
Taking the square root of both sides, we find that the initial velocity is approximately 9.64 m/s.
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