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

Answer :

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 + at

Given is a a pitcher threw a baseball straight up at 35.8 m/s

initial velocity [v] = 35.8 m/s

acceleration [g] = - 9.8 m/s²

time [t] = 2.5 seconds

Using first equation of motion, we get -

v = u + at

v = 35.8 - 9.8 x 2.5

v = 35.8 - 24.5

v = 11.3 m/s

Therefore, the velocity of the baseball after 2.5 seconds will be 11.3 m/s.

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Rewritten by : Barada

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