Final answer:
The hangtime for a ball launched with a velocity of 9.1 m/s is determined to be approximately 1.86 seconds, assuming it was launched vertically upwards, which is closest to option (a) 1.85 seconds.
Step-by-step explanation:
To determine how long a ball launched with a velocity of 9.1 m/s will stay in the air, we can use the kinematic equations for projectile motion under the influence of gravity. Since the question does not specify the angle at which the ball is launched, we'll assume it is launched vertically upwards. In this case, the time for the ball to reach its peak is given by the time it takes for the upward velocity to be reduced to zero due to gravity (9.8 m/s²). The formula to calculate this is t = v / g, where t is the time, v is the initial velocity, and g is the acceleration due to gravity.
t = 9.1 m/s / 9.8 m/s²
This gives the time to reach the peak of the trajectory. To find the total time in the air (hangtime), we double this time to account for the time taken to go up and then come back down.
Total hangtime = 2 * (9.1 m/s / 9.8 m/s²)
After calculation, the hangtime is approximately 1.86 seconds, which is closest to option (a) 1.85 seconds.