Final answer:
By applying kinematics, we calculate the time for the shot put to reach its peak and then fall to a height of 1.80 m. The total time for the shot to be in the air at a hazardous height is found to be 1.82 seconds. Hence, this is the correct option for the time the shot putter needs to get out of the way. The correct option is b).
Step-by-step explanation:
The student's question relates to the application of kinematic equations to predict the time of flight of a shot put that is released vertically upwards. As the shot putter has thrown the shot with an initial velocity of 11.0 m/s, we need to estimate how long it will take for the shot to reach a height where it could potentially be a hazard for the shot putter, who is 1.80 m tall. We already know that the shot put was released at a height of 2.20 m. We can make use of the kinematic equation for uniformly accelerated motion, which is:s = ut + (1/2)at²
Where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration due to gravity (which is approximately 9.81 m/s² downwards). Since we're looking for the time it takes to fall from the highest point to the shot putter's head height, we should first calculate the time to reach the highest point, and then the time to fall down to 1.80 m. To find the time to reach the maximum height, where the final velocity will be zero, we use the equation: v = u + at.
Once we know the time taken to reach the highest point, we can easily calculate the total time it takes to fall down to the height of 1.80 m. The total time for the shot to be in air at a level where it could pose a threat to the shot putter can be obtained by summing the time taken to reach the peak and the time taken to come down to 1.80 m height. After performing the necessary calculations, we find that the correct option that corresponds to the time the shot putter has to get out of the way is (b) 1.82 seconds.