Question 38: Motion in 1D

Question

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

Accepted Answer

The juggler throws n balls per second.
$$	herefore$$ Interval between balls = $${1 \over n}$$ seconds
At maximum height velocity of first ball $$v = 0$$
Juggler throws all balls with same initial velocity $$ = u$$
For 1st ball,
$${u_1} = u$$, $${v_1} = 0$$, $$a =  - g$$, $$S = {H_{\max }}$$
Using formula,
$${v^2} = {u^2} + 2as$$
$$0 = {u^2} - 2g\,{H_{\max }}$$
$$ \Rightarrow {H_{\max }} = {{{u^2}} \over {2g}}$$ ...... (1)
And using formula,
$$v = u + at$$
$$ \Rightarrow 0 = u - gt$$
$$ \Rightarrow t = {u \over g}$$
$$	herefore$$ Time taken by ball 1 to reach maximum height $$({H_{\max }}) = {u \over g}$$
According to the question,
$$t = {1 \over n}$$
$$ \Rightarrow {u \over g} = {1 \over n}$$
$$ \Rightarrow u = {g \over n}$$ ....... (2)
Putting value of $$u$$ in equation (1), we get
$${H_{\max }} = {{{{\left( {{g \over n}} ight)}^2}} \over {2g}}$$
$$ = {{{g^2}} \over {2{n^2}g}}$$
$$ = {g \over {2{n^2}}}$$

Answer

g/2n

Answer

g/n

Answer

2gn

Answer

g/2n2

Motion in 1D

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

Choose the correct answer: