A basketball player achieves a hang time of 0.773 s in dunking the ball.
What vertical height will he attain? The acceleration of gravity is 9.8 m/s\(^2\).
Answer (3)
The answer is 2.93m (to 2 dp)
To calculate the vertical height a basketball player attains during a jump, we can use kinematic equations for projectile motion. Since the player achieves a hang time of 0.773 seconds, and the acceleration due to gravity is 9.8 m/s², we can first determine the time it takes for the player to reach the peak of the jump, which is half of the total hang time. The peak time is therefore 0.773 s / 2, which equals 0.3865 seconds.
Using the kinematic equation h = v₀t + 0.5gt², where h is the height, v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity, we can solve for the initial vertical velocity.
Since the player reaches maximum height at the peak, the velocity at the peak is 0 m/s.
The equation simplifies to h = -0.5gt² since v₀ = 0 m/s at the peak.
Therefore, the height is -0.5 x (-9.8 m/s²) x (0.3865 s)², which gives us a vertical height of approximately 1.85 meters.
The basketball player attains a vertical height of approximately 1.85 meters during a jump with a hang time of 0.773 seconds. This calculation considers the effects of gravity and the relationship between time and velocity. The player reaches this height during the ascent phase of the jump before descending back down.
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