Question

A certain juggler usually tosses balls vertically to a height H. To what height must they be tossed if they are to spend twice as much time in the air?

Short answer

The height to which the balls must be tossed if they are to spend twice as much time in the air is‘4H’.

Step-by-step solution

To understand the concept

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The displacement in kinematic equation is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2$

In this case x is taken as H

Step 2: Calculations of height

We have to find the relation between H and t by using the following formula

$\begin{array}{rcl}x& =& v_{0}t+\frac{1}{2}a{t}^2\\ H& =& 0\times t+\frac{1}{2}\times g\times t^2\\ H& =& \frac{1}{2}\times g\times t^2\end{array}$

Now, we have doubled the time. So, plug $2ts$and find$H\text{'}$

$H\text{'}=\frac{1}{2}\times g\times {\left(2t\right)}^2$

Taking the ratio of these two equations, we have

$\begin{array}{rcl}\frac{H\text{'}}{H}& =& \frac{4t^2}{t^2}\\ H\text{'}& =& 4H\end{array}$

So, at $4H$height, they must be tossed.