Question

A ball is thrown vertically upwards. It goes to a height of 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m/$s^2$ . Find the initial and final velocity of the ball and also find the time.

Answer 1

Step 1: Given data:

The ball is thrown vertically upwards,

The maximum height covered by the ball($h$) = $20m$

The acceleration due to gravity($g$) = $10m{s}^{-2}$

Step 2: Concept and formula used:

The first equation of motion under gravity is given by;

$v=u+gt$ (For moving vertically downward)

And,

$v=u+(-g)t$

$v=u-gt$

( When the body is thrown upwards.)

At maximum height,

The velocity of the body= $0$

In falling from the maximum height,

The initial velocity of the ball = $0$

The third equation of motion under gravity is given by

$v^2=u^2+2gh$ (when body is coming downward from a height)

And,

$v^2=u^2+2(-g)h$

$v^2=u^2-2gh$ ( Body is moving vertically upwards)

(Note:

1. The acceleration due to gravity is taken as negative when a body is moving upward.
   
2. When a body is falling towards the earth, the acceleration due to gravity is taken as positive.​)

Step 3: Calculation of initial velocity of the ball while moving upward:

Using the third equation of motion gravity;

$0=u^2-2\times 10m{s}^{-2}\times 20m$

$u^2=400$

$u=20m{s}^{-1}$

The initial velocity of the ball is $20m{s}^{-1}$ .

Step 4: Calculation of the final velocity of the ball while returning to the ground:

While returning to the ground, the ball covers the same height that is $20m$ , therefore again using the third equation of motion,

$v^2=0+2\times 10m{s}^{-2}\times 20m$

$v^2=400$

$v=20m{s}^{-1}$

Thus,

While returning the final velocity of the ball is $20m{s}^{-1}$ .

Step 5: Calculation of the total time taken by the ball in its total journey:

Let, $t_1$ be the time taken by the ball to reach to the maximum height,

Then using the first equation of motion under gravity

$0=20m{s}^{-1}-10m{s}^{-2}\times t_1$

$10\times t_1=20$

$t_1=\frac{20}{10}$

$t_1=2sec$

If the time taken by the ball to return to the ground from the maximum height be $t_2$ the,

Using the first equation of motion under gravity we get

$20m{s}^{-1}=0+10m{s}^{-2}\times t_2$

$10\times t_2=20$

$t_2=\frac{20}{10}$

$t_2=2sec$

Thus, the total time taken by the ball for its journey will be;

$t=t_1+t_2$

$t=2sec+2sec$

$t=4sec$

Hence,

The initial velocity of the ball(while moving vertically upward) = $20m{s}^{-1}$

The final velocity of the ball(while moving vertically downward = $20m{s}^{-1}$

Total time taken by the ball = $4sec$