Question

37. A ball is dropped from top of a tower 250 m high.
At the same time, another ball is thrown upwards
with a velocity of 50 ms-t from the ground. The
time at which they cross each other is
(1) 3 second
(2) 5 second
(3) 10 second
(4) 12 second​

Answer 1

Answer:

$\displaystyle{t=5 \ sec}$

2. option is correct .

Explanation:

Given :

Total height ( H ) = 250 m

Initial velocity ( u ) = 0 m / sec

Final velocity ( v ) = 50 t m / sec

We have to find time when they cross each other

Case first :

$\displaystyle{h=ut+1/2at^2}\\\\\\\displaystyle{h=0+1/2\times10\times t^2}\\\\\\\displaystyle{h=5t^2}$

Case second :

$\\\\\\\displaystyle{h'=50t-5t^2}\\\\\\\displaystyle{We \ have \ h+h'=250m}$

Now put the value here we get

$\displaystyle{h+h'=5t^2+50t-5t^2}\\\\\\\displaystyle{250=50t}\\\\\\\displaystyle{5t=25}\\\\\\\displaystyle{t=5 \ sec}$

Thus at 5 sec they will cross each other .

Answer 2

Answer:

The correct answer is Option => B (5 seconds).  

Explanation:

Given Problem:

A ball is dropped from top of a tower 250 m high.  At the same time, another ball is thrown upwards  with a velocity of 50 ms-t from the ground. The  time at which they cross each other is

(1) 3 seconds

(2) 5 seconds

(3) 10 seconds

(4) 12 second​s

Solution:

To find:

The time taken by them to cross each other.

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.

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.

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

= 5s

Hence,

Option B (5s) is the Answer.