Question

from the top of a tower 100m in height , a ball is drop and at same time another ball is projected vertically upwards from the ground with a velocity of 25m/sec. find where and when the two balls meet ( use g= 9.8m' sec )​

Answer 1

Let the ball dropped down is a and the ball thrown up is b. also the stone thrown upwards cover a distance of x and the other covers a distance of (100 –x).

For ball a

u=0

g=10m/s  

2

 

d=(100−x)

Using the equation

s=ut+  

2

1

​  

at  

2

 

100−x=5t  

2

.........(1)

For ball b

d=x

g=−10m/s  

2

 

u=25m/s

s=ut+  

2

1

​  

at  

2

 

x=25t−5t  

2

............(2)

Solving equation (1) and (2)

100=25t

t=4seconds

Put the value of t in equation (1)

x=100−80

x=20m

They will meet at distance of 80 m from the ground after t = 4 seconds

Answer 2

They will meet instinctively. Here's how.

Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.

Let the ball dropped down be b1. So the ball thrown up is b2.

b1 => d = 100-x

g = 9.8 m/s sq.

u = 0

We know, s= ut + 1/2at sq.

Putting values,

100-x = 4.9t sq. ....(1)

b2 => d = x

g = -9.8 m/s sq.

u = 25 m/s

We know, s= ut + 1/2at sq.

Putting values,

x = 25t -4.9t sq. ....(2)

Adding (1) and (2),

100-x +x = 4.9t sq. + 25t - 4.9t sq.

t = 4 secs.

Puuting t = 4 in (1),

100-x = 4.9t sq.

100-x =19.6

x = 80.4 m

So, they meet at 80.4 m from ground after 4 seconds.