Question

a pitcher threw a baseball straight up at 35.8 meters per second. what was the balls velocity after 2.50?

Answer 1

Answer:

$v=11.3 ms^{-1}$

Explanation:

According to question, a pitcher threw a baseball straight up at 35.8 meters per second.

Use the expression of the equation of motion.

v= u+at

Here, v is the final velocity.

Put $a= g= -9.8 ms^{-2}$, t= 2.50 s, $u=35.8 ms^{-1}$ and u = 0.

$v= 35.8+ (-9.8)\times 2.50$

$v=11.3 ms^{-1}$

Therefore, the velocity of ball after 2.50 s is $v=11.3 ms^{-1}$.