a pitcher threw a baseball straight up at 35.8 meters per second. what is the ball's velocity after 2.50 seconds?

The velocity of the baseball after 2.5 seconds will be 11.3 m/s.State first equation of motion?The first equation of motion is -v = u + atGiven is a a pitcher threw a baseball straight up at 35.8 m/sinitial velocity [v] = 35.8 m/sacceleration [g] = - 9.8 m/s²time [t] = 2.5 secondsUsing first equation of motion, we get -v = u + atv = 35.8 - 9.8 x 2.5v = 35.8 - 24.5v = 11.3 m/sTherefore, the velocity of the baseball after 2.5 seconds will be 11.3 m/s.To solve more questions on Kinematics, visit the link below-https://brainly.com/question/15319811#SPJ2

ok the velocity of an object in free fall is given by the equation :v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time. so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s