Question

A juggler throws the balls in the air. How high will each ball rise if he throws n balls per second?
A. $\dfrac{{4{{\rm{n}}^2}}}{g}$
B. $\dfrac{g}{{{\rm{2}}{{\rm{n}}^2}}}$
C. $\dfrac{g}{{4{\pi ^2}}}$
D. $\dfrac{{4g}}{{{{\rm{n}}^2}}}$

Answer 1

We will be using the equations of motion, mainly the first and third equation. These equations tell us about the relation between initial velocity, final velocity, acceleration and time period of a moving body.

Complete step by step answer:
First equation of motion of a body under the influence of gravity.
$v = u - gt$……(1)

Here v is initial velocity, u is final velocity and t is time period.

Third equation of motion of a body under the influence of gravity.
${v^2} = {u^2} - 2gs$……(2)

Here s is the distance travelled. When a ball is thrown upward, at the instant of maximum height attained its velocity is zero. V=0 m/s Here v is the final velocity of a ball from a number of balls thrown upward by the juggler. It is given that the juggler throws n number of balls in one second so the time taken to throw one ball will be given as $$\left( {\dfrac{1}{{\rm{n}}}} \right)$$ second. Substitute $$\dfrac{1}{{\rm{n}}}$$ for t and $$0$$ for v in equation (1). $$\begin{array}{l} v = u - g\left( {\dfrac{1}{{\rm{n}}}} \right)\\\ u = \dfrac{g}{{\rm{n}}} \end{array}$$ Let H is the maximum height attained by each ball. Substitute H for s, $$0$$ for v and $$\dfrac{g}{{\rm{n}}}$$ for u in equation (2). $$\begin{array}{l} {0^2} = {\left( {\dfrac{g}{{\rm{n}}}} \right)^2} - 2gH\\\ H = \dfrac{g}{{2{{\rm{n}}^2}}} \end{array}$$ Therefore, maximum height attained by each ball if juggler throws n balls per second is given by $$\left( {\dfrac{g}{{2{{\rm{n}}^2}}}} \right)$$ **So, the correct answer is “Option B”.** **Note:** Take extra care while using the equations of motion because under the influence of gravity their forms have slight changes than conventional ones.