Question

A ball is pushed with an initial velocity of 4.0 m/s. The ball rolls down a hill with a constant acceleration of 1.6 m/s2. The ball reaches the bottom of the hill in 8.0 s. What is the ball's velocity at the bottom of the hill?

Answer 1

Answer:

16.8 m/s

Explanation:

As the ball is subjected to a constant acceleration of 1.6m/s2 in 8 seconds before reaching the bottom of the hill, its increment in velocity must be

$\Delta v = a\Delta t = 1.6 * 8 = 12.8 m/s$

Since it was pushed with an initial velocity of 4m/s, its final velocity when reaching bottom of the hill is

$v = v_0 + \Delta v = 4 + 12.8 = 16.8 m/s$

Answer 2

The ball's velocity will be "16.8 m/s".

Acceleration and Velocity:

According to the question,

Initial velocity = 4.0 m/s

Acceleration = 1.6 m/s²

Time = 8.0 seconds

The increment in velocity be:

→ $\Delta v = a \Delta t$

By substituting the values,

        $= 1.6\times 8$

        $= 12.8 \ m/s$

hence,

The ball's final velocity be:

→ $v = v_0+ \Delta v$

     $= 4+12.8$

     $= 16.8 \ m/s$

Thus the above answer is correct.

Find out more information about acceleration here:

https://brainly.com/question/460763