Question

A ball of mass m = 1.9 kg is released from a height of h = 2.9 m into a basin of water. At a time of t = 1.33 s after striking the surface of the water, the ball's velocity has decreased by 50%. What is the final velocity of the ball at t = 1.33 s? A) 4.5 m/s B) 7.25 m/s C) 3.75 m/s D) 2.25 m/s

Answer 1

The Mechanics initial velocity of the ball when it hits the water is calculated to be 7.57 m/s. Given that this speed is reduced by 50% in 1.33 seconds, the final velocity at this time is 3.785 m/s. The answer closest to this in the provided options is C) 3.75 m/s.

Step-by-step explanation:

The subject of this question is physics. Specifically, it concerns the principles of mechanics, including gravity, acceleration, and velocity. To determine the final velocity of the ball 1.33 seconds after striking the water's surface, first, we need to calculate the initial velocity when the ball strikes the water.

Using the physics formula for velocity in free-fall (v = sqrt(2gh)), we find the initial velocity to be approximately sqrt(2 * 9.8 m/s² * 2.9m) = 7.57 m/s. Then, following the information given in the problem, the ball's velocity has decreased by 50% 1.33 seconds after hitting the water. So, the final velocity of the ball is 50% of the initial velocity, which is 0.5 * 7.57 m/s = 3.785 m/s. This is closest to option C) 3.75 m/s, which should be the correct answer.

Learn more about Mechanics