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Question: A girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of \[50{\tex...

A girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of 50 ms−1.50{\text{ }}m{s^{ - 1}}. If the platform starts moving up at 5 ms−15{\text{ }}m{s^{ - 1}} and the girl again throws the similar ball in similar way. The time taken in the previous case and the time taken in the second case to return to her hands are
A. 10s, 5s10s,{\text{ }}5s
B. 5s, 10s5s,{\text{ }}10s
C. 10s, 10s10s,{\text{ }}10s
D. 5s, 5s5s,{\text{ }}5s

Explanation

Solution

In the question, we are given two cases, where a girl throws a ball in two different conditions, we need to find the taken in both cases. So, for that we will take the time to reach the highest point and come back, similarly in second case, for that we will put the formula, t=ugt = \dfrac{u}{g}, then on putting the value in the formula, we will get our required answer.

Complete step by step answer:
We have been given that a girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of 50 ms−1.50{\text{ }}m{s^{ - 1}}. It is given that if the platform starts moving up at 5 ms−15{\text{ }}m{s^{ - 1}} and the girl again throws the similar ball in similar way. We need to find the time taken in the first case and the time taken in the second case for the ball to return to her hands.
The given maximum possible speed of throwing ball upwards in first case = 50 ms−1 = {\text{ }}50{\text{ }}m{s^{ - 1}}
First, we will evaluate the time taken by ball to reach the highest point, for that we will use the formula mentioned below.
Time taken to reach the highest point, t=ugt = \dfrac{u}{g}
where, u == maximum possible speed (50 ms−1)\left( {50{\text{ }}m{s^{ - 1}}} \right)
g == acceleration due to gravity (10 ms−1)\left( {10{\text{ }}m{s^{ - 1}}} \right)
Now, on putting the values, in the above formula, we get
t=5010t = \dfrac{{50}}{{10}}
​= 5sec = {\text{ }}5sec
Now the total time taken by ball to come back = 5 + 5 = 10sec = {\text{ }}5{\text{ }} + {\text{ }}5{\text{ }} = {\text{ }}10sec
In the second case, it is given that the platform starts moving up at 5 ms−15{\text{ }}m{s^{ - 1}} in this situation, the acceleration is zero.
So, the motion of the girl and the ball with respect to each other will be unaffected.
So, the total time taken in second case will be just like the first case, i.e., 10sec.10sec.
Therefore, the time taken in the first case and the time taken in the second case for the ball to return to her hands are 1010 seconds each.

So, the correct answer is “Option C”.

Note:
In the question, it was mentioned that the platform starts moving up at 5 ms−1,5{\text{ }}m{s^{ - 1}}, but in the solution, we have not taken this value, because when the platform move at a uniform speed, that time the acceleration is zero, and the motion is also unaffected. Hence, this value was not considerable