Two balls A and B are thrown with same speed u from the top of a tower .Ball A is thrown vertivally upwards and ball B is thrown vertically downwards.If time taken taken by the ball A and ball B to reach the ground are 6 and 2 sec Respectively.

Find the distance travelled by ball in t=3 s.Ans=25 m

Find the height of the tower.Ans=60 m

Let the height of the tower be ‘h’.

Ball A is thrown vertically upwards with speed ‘u’. It takes 6 s to reach the ground.

Ball B is thrown vertically downwards with speed ‘u’. It takes 2 s to reach the ground.

Now for ball A,

-h = ut – ½ gt² [h and g are downwards so negative signs are added]

=> -h = 6u – ½ × 10 × 6²

=> -h = 6u – 180 …………….(1)

And for ball B,

-h = -ut - ½ gt² [here u, h and g are all downwards]

=> h = 2u + ½ × 10 × 2²

=> h = 2u + 20 …………………(2)

From (1) and (2) we have,

h = 60 m and u = 20 m/s

Now, lets find the distance travelled by A in 3 s.

For that lets first find out the maximum height reached by the ball A using,

v² = u² + 2ax

=> 0 = 20² – 2 × 10x

=> x = 20 m

Time taken in travelling this 20 m is found using,

v = u + at

=> 0 = 20 – 10t

=> t = 2 s

Now, remaining 1 s it free falls from top, so distance travelled in this 1 s is,

x₂ = 0 + ½ gt²

=> x₂ = ½ × 10 × 1²

=> x₂ = 5 m

So, the distance traveled in 3 s by A is = 20 + 5 = 25 m