Question

When a pitcher throws a baseball, it reaches a top speed of 45 m/s. If the
baseball takes 0.9 seconds to travel from the pitcher to the catcher, what is
its acceleration? (Assume the ball is moving at 0 m/s right before it leaves
the pitcher's hand.)



A. 45 m/s2

B. 50 m/s2


C. 55m/s 2


D. 40 m/s2

Answer 1

Answer:

B. 50 m/s²

Explanation:

Given:

Initial velocity of the ball is, $u=0\ m/s$

Final velocity of the ball is, $v=45\ m/s$

Time taken by the  ball is, $t=0.9\ s$

Acceleration is defined as the rate of change of velocity. So, the velocity of the ball is changing with time. So, acceleration is given as:

$a=\frac{v-u}{t}$

Plug in 0 for 'u', 45 for 'v', 0.9 for 't' and solve for acceleration, 'a'. This gives,

$a=\frac{45-0}{0.9}\\\\a=\frac{45}{0.9}\\\\a=50\ m/s^2$

Therefore, the acceleration of the ball is 50 m/s².