1.

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

Answer»

\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.


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