Question

A juggler performs in a room whose ceiling is 3m above the level of his hands. He throws a ball vertically upwards so that it just reaches the ceiling.
$(a)$ What time is required for the ball to reach the ceiling?

Answer 1

- Hint – Let the time required be (t) sec, the initial velocity be u m/s and the final velocity be v m/sec. As the ball reaches the ceiling the final velocity of the ball becomes zero and the acceleration due to gravity will be downwards so it’s sign will be negative. Use this concept along with the first and the third equation of motion to get time.

Formula used – $v = u + at$, ${v^2} = {u^2} + 2as$
Let the initial velocity of the ball be (u) m/s.
And the final velocity of the ball be (v) m/s.

Complete step-by-step solution -

The distance (s) travelled by the ball = 3 m.
Let the time taken to reach the ceiling = t sec.
Now from first equation of motion which is given as,
$v = u + at$...................... (1)
Where v = final velocity, u = initial velocity, a = acceleration and t = time taken.
Now when the ball is thrown vertically upwards so the ball has some initial velocity but when the ball reaches at the ceiling the final velocity becomes zero and the acceleration due to gravity is acting downwards on the ball.
As we know acceleration (a) due to gravity = 9.8 $m/{s^2}$.
$ \Rightarrow v = 0m/s$ and $a = - 9.8m/{s^2}$(‘-’ sign is because it is acting downwards).
So substitute these values in equation (1) we have,
$ \Rightarrow 0 = u + \left( { - 9.8} \right)t$
$ \Rightarrow u = 9.8t$................... (2)
Now from third equation of motion we have,
${v^2} = {u^2} + 2as$
Now substitute the values in this equation we have,
$ \Rightarrow {0^2} = {\left( {9.8t} \right)^2} + 2\left( { - 9.8} \right)\left( 3 \right)$
Now simplify the equation we have,
$ \Rightarrow {\left( {9.8t} \right)^2} = 2\left( {9.8} \right)\left( 3 \right)$
\[ \Rightarrow {t^2} = \dfrac{{2\left( {9.8} \right)\left( 3 \right)}}{{{{\left( {9.8} \right)}^2}}} = \dfrac{6}{{9.8}} = \dfrac{3}{{4.9}} = \dfrac{{30}}{{49}}\]
Now take square root on both sides we have,
\[ \Rightarrow t = \sqrt {\dfrac{{30}}{{49}}} = \dfrac{{\sqrt {30} }}{7} = 0.7824\] Seconds.
So this is the required time taken by the ball to reach the ceiling.

Note – Whenever we face such types of questions the key concept here is that the final velocity of the ball is zero so using this and from the first equation of motion v = u + at find out its initial velocity in terms of t, i.e. time. Then substitute this value in the third equation of motion as above and simplify, we will get the required time taken by the ball to reach the ceiling, remember that the value of acceleration = -9.8 $m/{s^2}$ (acting downward).