Question

During the 1990 Olympics, an all-time record shot put was made with a distance of 23.12 meters (75 ft 10.2 in.). If the athlete released the shot at a height of 2.25 meters with a velocity of 14.3 m/s at an angle of 35.0° to the ground, what was its vector velocity when it impacted the ground at the end of its flight?

A) 14.3 m/s at 35.0°
B) 12.8 m/s at 40.0°
C) 11.6 m/s at 32.0°
D) 13.2 m/s at 28.0°

Answer 1

The question asks for the vector velocity of a shot put at impact during an Olympic record set in 1990, which requires an application of projectile motion principles in physics. Due to a lack of relevant computational data, the answer can't be provided with certainty.

Step-by-step explanation:

The question concerns a scenario from the 1990 Olympics where a shot put record was made, and it asks for the vector velocity of the shot at the moment of impact with the ground, given that it was released at a certain height and velocity and at a specific angle. This is a classic problem in projectile motion, a part of physics that deals with objects moving under the influence of gravity after being projected into the air.

To determine the correct answer, one would have to consider the two components of the shot's motion separately—the horizontal and vertical components. However, due to the complexity of the question and the absence of a possible computational resolution within the provided information, it's not possible to accurately determine the velocity at impact. We need either the time of flight or additional data regarding the behavior of the shot as it travels through the air.