Question

The velocity of a ball is thrown up is 20 meters/second. It attains zero velocity after 3.5 seconds. What is the average acceleration of the ball during its upward flight?

Answer 1

Average acceleration: ( v - v o ) / Δ t
v = 0 m/s,  v o = 20 m/s,  Δ t = 3.5 s
a (average) = ( 0 m/s - 20 m/s )/3.5 s = - 20 m/s / 3.5 s =
 = - 5.7142857 m/s² ≈ - 5.71 m/s²

Answer 2

Answer:

Acceleration = 5.714 meter per second²

Step-by-step explanation:

A ball has been thrown up with a speed of 20 meters per second.

After 3.5 seconds speed of the ball become zero that means ball attains it's maximum height after 3.5 seconds.

Now we have to calculate the average acceleration during the upward motion.

Since, $\text{Acceleration}=\frac{\text{Change in velocity}}{\text{Duration of motion}}$

Acceleration = $\frac{20-0}{3.5}$

                     = 5.714 meter per second²