Question

a boy throws a ball upwards at a speed of 20 m/s . What is the speed when it returns to his hand? How long was it in air​

Answer 1

Given:

A boy throws a ball upwards at a speed of 20 m/s .

To Find:

What is the speed when it returns to his hand? How long was it in air​?

Solution:

This can be solved by equation of motion:

Before apply equation of motion, let us take convection sign:

Upward taking as positive sign

Downward taking as negative sign

$\mathbf{\textrm{Acceleration due to gravity }=a=g=-10 \ \dfrac{m}{s^{2}}}$    (because it direction is downward)

$\mathbf{\textrm{Initial velocity (u)}=+20 \ \dfrac{m}{s}}$          (because it direction is upward)

Total displacement from starting of throwing ball to catch ball:

Total displacement (s) = 0 m

From equation of motion:

$\mathbf{v^{2}=u^{2}+2\times a\times s}$

On putting respecting value:

$\mathbf{v^{2}=20^{2}+2\times 10\times 0}$

v = ±20

So direction of final velocity is in downward direction:

$\mathbf{v=-20 \ \dfrac{m}{s}}$

For duration of time of ball in air  , we again use equation of motion taking from initial to final period:

$\mathbf{v=u+at}$

$\mathbf{-20=20-10\times t}$

On solving we get:

Time of duration of ball in air = t =  4 second

Means ball return to hand with same speed of 20 m/s and it is 4 second in air.