Question

A juggler performs in a room whose ceiling is $3\,{\text{m}}$ above the level of his hands. He throws a ball vertically upwards so that it just reaches the ceiling. (a) With what initial velocity does he throw the ball?

Answer 1

Use the kinematic equation for final velocity of an object. This kinematic equation gives the relation between the final velocity of the object, initial velocity of the object, acceleration of the object and displacement of the object. Rewrite this equation for the vertical motion of the ball and calculate the initial vertical velocity of the ball.

Formula used:
The kinematic equation for final velocity $v$ of an object in terms of its displacement is given by
${v^2} = {u^2} + 2as$ …… (1)
Here, $u$ is initial velocity of the object, $a$ is acceleration of the object and $s$ is displacement of the object.

Complete step by step answer:
We have given that the height of the ceiling of the room from the level of the hands of the juggler is $3\,{\text{m}}$.
$h = 3\,{\text{m}}$
We have asked to calculate the initial velocity with which the juggler throws the ball vertically towards the ceiling.We can calculate the velocity with which the juggler throws the ball towards the ceiling using equation (1).
Rewrite equation (1) for the ball in the free fall.
$v_y^2 = u_y^2 - 2gh$ …… (2)
Here, ${v_y}$ is the final vertical velocity of the ball when it touches the ceiling and ${u_y}$ is the initial vertical velocity of the ball.

When the ball touches the ceiling, its velocity becomes zero.
${v_y} = 0\,{\text{m/s}}$
Rearrange equation (2) for ${u_y}$.
${u_y} = \sqrt {v_y^2 + 2gh}$
Substitute $0\,{\text{m/s}}$ for ${v_y}$, $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $3\,{\text{m}}$ for $h$ in the above equation.
${u_y} = \sqrt {{{\left( {0\,{\text{m/s}}} \right)}^2} + 2\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {3\,{\text{m}}} \right)}$
$\Rightarrow {u_y} = \sqrt {58.8}$
$\therefore {u_y} = 7.66\,{\text{m/s}}$

Hence, the initial velocity with which the juggler throws the ball towards the ceiling is $7.66\,{\text{m/s}}$.

Note: The students may think that why we have changed the positive sign in the kinematic equation to negative sign. The students should keep in mind that when the ball is in the vertical motion towards the ceiling of the room, the acceleration of the ball is equal to the acceleration due to gravity which always acts in the downward direction. Hence, the negative sign of the acceleration due to gravity is used.