Question

A ball is thrown vertically upwards from the 12m level with an initial velocity of 18 m/s. At the same instant an open platform elevator passes the 5m level, moving upwards with a constant velocity of 2m/s. Determine (g = 9.8 $m/{s^2}$)
(a) When and where the ball will meet the elevator?
(b) The relative velocity of the ball with respect to the elevator when the ball hits the elevator.

Answer 1

Hint: In this question consider the Initial velocity of ball u = 18 m/s, initial level of the ball from the ground ${y_b}$= 12 m, velocity of elevator ${v_e}$= 2 m/s (constant velocity, initial level of elevator from the ground ${y_e} = 5$ m. Then use a second equation of motion to get the distance travelled by the ball and the elevator. The distance travelled by the elevator must be equal to the distance travelled by the ball to get the first part. For the second part use the concept that the relative velocity of ball is the difference of final velocity of ball and velocity of elevator.
Formula used – $v = u + at$, $s = ut + \dfrac{1}{2}a{t^2}$

Complete Step-by-Step solution:
Given data:
Initial velocity of ball u = 18 m/s.
Initial level of the ball from the ground ${y_b}$= 12 m
Velocity of elevator ${v_e}$= 2 m/s (constant velocity)
Initial level of elevator from the ground ${y_e} = 5$ m
Now from first equation of motion we have,
$v = u + at$ m/s.
Where v = final velocity, u = initial velocity, a = acceleration and t = time.
As the ball is going upward so acceleration is acting downwards therefore a = - 9.8 $m/{s^2}$.
Let the final velocity of ball = ${v_b}$
Therefore,
$ \Rightarrow {v_b} = 18 - 9.8t$
Now the second equation of motion is given as
$s = ut + \dfrac{1}{2}a{t^2}$
So the distance covered by the ball if the initial distance of the ball from the round is 12 m.
$ \Rightarrow {d_b} = 12 + 18t + \dfrac{1}{2}\left( { - 9.8} \right){t^2}$
And the distance covered by the elevator if the initial distance of the elevator from the ground is 5 m.
As the elevator travels with the constant velocity so the acceleration of the elevator is zero (0).
$ \Rightarrow {d_e} = 5 + 2t + \dfrac{1}{2}\left( 0 \right){t^2} = 5 + 2t$
Now the ball will meet the elevator if both the distances are equal.
$ \Rightarrow {d_b} = {d_e}$
$ \Rightarrow 12 + 18t + \dfrac{1}{2}\left( { - 9.8} \right){t^2} = 5 + 2t$
$ \Rightarrow 4.9{t^2} - 16t - 7 = 0$
Now apply quadratic formula we have,
$ \Rightarrow t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where [a = 4.9, b = -16 and c = -7]
$ \Rightarrow t = \dfrac{{16 \pm \sqrt {{{\left( {16} \right)}^2} - 4\left( {4.9} \right)\left( { - 7} \right)} }}{{2\left( {4.9} \right)}} = 3.656, - 0.391$
Negative time is not possible.
$\left( a \right)$ So after 3.656 sec ball and elevator will meet.
And at distance ${d_e} = {d_b} = 5 + 2\left( {3.656} \right) = 12.312$ meter.
$\left( b \right)$ Now the relative velocity of the ball is the difference of the final velocity of the ball and velocity of the elevator.
$ \Rightarrow {v_{b|e}} = {v_b} - {v_e}$
$ \Rightarrow {v_{b|e}} = 18 - 9.8t - 2 = 16 - 9.8\left( {3.656} \right) = - 19.8288$ m/s.
Here negative sign indicates that the ball is going downwards w.r.t. the elevator (i.e. final velocity of the ball is less than the velocity of the elevator).
So this is the required answer.

Note – The trick point here was that the final velocity of ball is taken out using the first equation of motion which is $v = u + at$, however we have not used it for the elevator as in the question it is given that the elevator is moving upwards with a constant velocity so the final velocity will not get changed as it will remain constant throughout for elevator.