Question

When a ball is thrown vertically upwards, it goes
through a distance of 19.6 m. Find the initial velocity of the ball and the time
laken by it to rise to the highest point. (Acceleration due to gravity,
= 9.8 m/s2)​

Answer 1

$\large \bf{ \color{magenta}{proper \: Question}}$

When a ball is thrown vertically upwards, it goes

through a distance of 19.6 m. Find the initial velocity of the ball and the time Taken by it to rise to the highest point. (Acceleration due to gravity,

= 9.8 m/s²)

$\large \bf{ \color{magenta}{Solution : }}$

Here , the ball is going up against the attraction of Earth, so it's velocity is decreasing continuosly . In other words we can say that the ball is being retarded . Thus the acceleration in the ball is negative which means that the value of g is to be used here with negative sign.

Here,

$\implies\rm{ \color{blue}{initial \: velocity \: of \: ball \: (u) =?}}$

$\implies \rm{ \color{blue}{final \: velocity \: (v) = 0}}$

$\implies \rm{ \color{blue}{acceleration \: due \: to \: gravity \: (g) = - 9.8 \: m/ {s}^{2} }}$

$\implies \rm{ \color{blue}{and\: height \: (h) = 19.6 \: m}}$

Now , putting all these values in the formula :-

$\implies \bf{ \color{red}{ {v}^{2} = {u}^{2} + 2gh}}$

We get ,

$\implies\rm \bold{ \color{darkblue}{ {(0)}^{2} = {u}^{2} + 2 \times ( - 9.8) \times 19.6}}$

$\implies \rm \bold{ \color{darkblue}{0 = {u}^{2} + 2 \times ( - 9.8) \times 19.6}}$

$\implies \rm \bold{ \color{darkblue}{0 = {u}^{2} - 19.6 \times 19.6}}$

$\implies \rm \bold{ \color{darkblue}{ {u}^{2} = {(19.6)}^{2} }}$

$\implies \rm \bold {\underline{ \color{magenta}{u = 19.6 \: m/s}}}$

Thus the velocity of the ball is 19.6 m/s which means that the ball has been thrown upwards With a velocity of 1o.6 m/s

Let us now calculate the time taken by the ball to reach the highest point.

Now , the initial velocity, the final velocity and the acceleration due to gravity, so the time taken can be calculated by using the equation :

$\implies \bf{ \color{red}{v = u + gt}}$

• Final velocity = o
• Initial velocity = 19.6 m/s
• Acceleration due to gravity = -9.8m/s²
• Time = ?

So putting the values in the above equation.

we get,

$\implies \rm \bold{ \color{darkblue}{0 = 19.6 + ( - 9.8) \times t}}$

$\implies \rm \bold{ \color{darkblue}{0 = 19.6 - 9.8t}}$

$\implies \rm \bold{ \color{darkblue}{9.8 \: t = 19.6}}$

$\implies \rm \underline \bold{ \color{darkblue}{t = 2s}}$

Thus the ball takes 2 seconds to reach the highest point of its upward journey.

______________________________

Note :-

• The ball will take an equal time that is 2 seconds to fall back to the ground.
• in other words , The ball will take a total of 2 + 2 = 4 seconds to reach back to the thrower.

Answer 2

Here , the ball is going up against the attraction of Earth, so it's velocity is decreasing continuosly . In other words we can say that the ball is being retarded . Thus the acceleration in the ball is negative which means that the value of g is to be used here with negative sign.