Question

A boy is throwing balls into the air one by one in such a way that when the first ball thrown reaches maximum height he starts to throw the second ball. He repeats this activity. To what height do the balls rise if he throws twice in a second?

Answer 1

We are given that time for two throws to be equal to 1 second. The balls are rising against gravity. We can study their motion using the equations of motion. By solving them, we shall be able get the height to which the balls rise.

Formula used: The equations of motion are given as:
$v = u + at \\\ S = ut + \dfrac{1}{2}a{t^2} \\\ {v^2} - {u^2} = 2aS \\\$

Complete step by step answer:
We are given that a boy is throwing balls into the air one by one in such a way that when the first ball thrown reaches maximum height he starts to throw the second ball. He repeats this activity. We need to find out the height to which the balls rise if he throws twice in a second.
Therefore, we have $t = 1s$. The final velocity of the balls must be zero when they reach their maximum height. Therefore, $v = 0$.
Now using these values in the first equation of motion, we obtain that
$v = u + at \\\ 0 = u + \left( { - g} \right) \times 1 \\\ u = g \\\$
Here v is the final velocity of the ball. Since the ball is being thrown upwards against
Now we shall use this information in the third equation of motion in the following way.
${v^2} - {u^2} = 2aS \\\ 0 - {g^2} = 2\left( { - g} \right)H \\\ 2gH = {g^2} \\\ H = \dfrac{g}{2} \\\$
If we take the value of acceleration due to gravity as $10m/{s^2}$, then maximum height reached by one ball is equal to
$H = \dfrac{{10}}{2} = 5m$
This is the required solution.

Note: It should be noted that when a body moves against the gravity, it decelerates and the acceleration is taken as negative. But when the body falls in the gravity then it accelerates and the acceleration is taken as positive. The balls decelerate as they travel upwards eventually their velocity reduces to zero and they start falling down accelerating in gravity.