Question

A juggler maintains four balls in motion, making each of them to rise a height of $20\;{\text{m}}$ from his hand. What time interval should he maintain, for the proper distance between them? ($g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}}$ )
(A) $3\;{\text{s}}$
(B) $\dfrac{3}{2}$
(C) $1\;{\text{s}}$
(D) $2\;{\text{s}}$

Answer 1

Hint: The equations of motion are relating the variables such as time, initial and final velocities, acceleration and the displacement of the system of motion. These equations of motion can be used for finding the unknown variables.

Useful formula The velocity of each ball and the time taken by each ball can be calculated from the equations of motion.
${v^2} - {u^2} = 2as$
And $v = u + at$
Where, $v$ is the final velocity, $u$ is the initial velocity, $s$ is the displacement ,$a$ is the acceleration and $t$ is time taken.

Complete step by step solution
the given data from the problem is;
Each ball has to rise a height of $20\;{\text{m}}$.
Consider each ball. To find the velocity of each ball when it reaches the height of $20\;{\text{m}}$, apply the equation of motion.
${v^2} - {u^2} = 2as$
When the ball reaches the hands of the Juggler then the velocity becomes zero. Hence the action is against the gravity the acceleration due to gravity is taken as negative.
Therefore substituting the values in the equation we get,
$
  0 - {u^2} = 2\left( { - 10\;{\text{m/}}{{\text{s}}^{\text{2}}}} \right) \times 20\;{\text{m}} \\
   - {u^2} = - 400\;{{\text{m}}^2}{\text{/}}{{\text{s}}^{\text{2}}} \\
  u = 20\;{\text{m/s}} \\
 $
Each ball has a velocity $20\;{\text{m/s}}$when it reaches the height of $20\;{\text{m}}$.
To find the time by which it reaches the height is calculated by another equation of motion.
$v = u + at$
Substituting the values in above expression,
$
  0 = 20\;{\text{m/s - 10}}\;{\text{m/s}} \times t \\
  t = 2\;{\text{s}} \\
 $
The ball has to reach the height and returns to the hands. Therefore the total time for each ball is
${\text{2}}\;{\text{s + 2}}\;{\text{s = 4}}\;{\text{s}}$
There are four balls. So, each ball must be thrown at an interval of $\dfrac{4}{4} = 1\;{\text{s}}$ .
For maintaining proper distance, the Juggler must throw each ball at an interval of $1\;{\text{s}}$ .

Thus, the answer is option C.

Note: When the action is opposite to the gravity, then the acceleration due to gravity is taken to be negative. Hence in the equation${v^2} - {u^2} = 2as$, instead of $a$ ,$\left( { - g} \right)$ is used. And for the body in rest, the velocity will be equal to zero.