Chapter 5, Problem 11A

Neglecting air resistance, if you throw a ball straight up with a speed of 20 m/s, how fast will it be moving when you catch it?

To determine

Speed of ball when it catches by a person.

Answer to Problem 11A

Speed of ball when it is caught by a person is $v=19.9\frac{\text{m}}{\text{s}}$ .

Explanation of Solution

Given:

Initial speed of ball, $v_i=20\frac{\text{m}}{\text{s}}$

Speed of ball at highest point, $v_h=0\frac{\text{m}}{\text{s}}$

Acceleration due to gravity, $g=9.\text{8}\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Formula used:

Equation of kinematics is,

  $v^2-v_0^2=2ad$

Where,

  $v$ = Final speed of an object

  $v_0$ = Initial speed of an object

  $a$ =Acceleration of an object

  $d$ = Distance travelled by an object

Calculation:

Maximum height of ball,

  $v^2-v_0^2=-2gh$

Negative sign of acceleration due to gravity indicates that, directions of velocity and acceleration due gravity are opposite.

Now, substitute known values in above equation,

  ${(0)}^2-{(20)}^2=-2\times 9.8\times h$

  $400=19.6\times h$

  $h=20.2$ m

Now, final speed of ball,

Substitute known values in above equation,

  $v^2-0=2\times 9.8\times 20.2$

  $v^2=395.92$

  $v=19.9\frac{\text{m}}{\text{s}}$

Conclusion:

Speed of ball when it is caught by a person is $v=19.9\frac{\text{m}}{\text{s}}$