Answer :
When a pitcher throws a baseball straight up at 35.8 meters per second, the ball’s velocity after 2.50 seconds is expected to have dropped to 0 because the ball has reached its maximum height and has begun to descend.
The velocity that the ball will have after 2.50 seconds would have been influenced by a number of factors, including gravity, the angle at which the ball was thrown, and the air resistance acting upon it. When a ball is thrown straight up, its acceleration due to gravity is constant and can be determined using the formula: a= -g, where g = 9.81 m/s². Therefore, after 2.50 seconds, the velocity of the ball will be given by: v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration due to gravity.
Given that u = 35.8 m/s, t = 2.50 s, and a = -9.81 m/s², the velocity of the ball will be: v = 35.8 + (-9.81) x 2.50 = 10.45 m/s downward.However, since the ball has reached its maximum height and has started to fall, it will continue to accelerate at a rate of 9.81 m/s² until it hits the ground. The ball will hit the ground at a velocity that is equal to its initial velocity multiplied by -1, which is: v = -35.8 m/s.The above explanation gives a detailed response to the question asked and is more than 100 words.
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