Question

A Ball is dropped from a top of a tower 250m high. At the same time, another ball is thrown upwards with a velocity of 50m/s . the time at which they cross each other is........options are....A 3sec.....B 5sec.....C 10 sec.....D 12sec

Answer 1

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Answer 2

The correct answer is option B (5 seconds).  

Given:

The initial velocity of the ball $\left(u_{1}\right)=0$

Final velocity$\left(u_{2}\right)=50 \mathrm{m} / \mathrm{s}$

Ball dropped from the height=250m

To find:

The time taken by them to cross each other

Solution:

We have  

$S_{1}+S_{2}=250 \text { and } t_{1}=t_{2}=t$

[This is when the ball crosses at each other]

$S_{1}=\frac{1}{2} g t^{2}=5 t^{2}$

$S_{2}=u_{2} t-\frac{1}{2} g^{2}$

$=50 t-5 t^{2}$

We have  

$S_1+S_2=250$

Substituting the values of $S_1 and S_2$

We get,

$5t^{2}+\left(50 t-5 t^{2}\right)=250$

50t=250

To find the value of t let us shift 50 to other side and write it in the denominator

$t=\frac{250}{50}$

=5s

Hence option B 5s is the solution.