Chapter 3, Problem 75A

To determine

To Find: The velocity of the ball.

Explanation of Solution

Given info:

Distance, $y=-2.5\text{m}$

Initial velocity, $v_0=2.\text{0}\text{m/s}$

Acceleration due to gravity, $g=-9.8{\text{m/s}}^{\text{2}}$

Formula used:

Kinematics third equation is:

  $v^2=v_0^2+2gy$

Here

  $v=$ Final velocity

  $v_0=$ Initial velocity

  $g=$ acceleration due to gravity

  $y=$ Height of object

Calculation:

Substituting the given values

  $v^2=v_0^2+2gy$

The Final velocity of the ball is

  $v=\sqrt{v_0^2+2gy}$

  $v=\sqrt{{\text{(2}\text{.0}\text{m/s)}}^2\text{+2(-9}\text{.8}{\text{m/s}}^{\text{2}}\text{)(-2}\text{.5}\text{m)}}$

  $v=7.3\text{m/s}$

Conclusion:

So, the ball covers trajectory pathways when throw from window and reached and hits the sidewalk at velocity $v=7.3\text{m/s}\text{.}$