Question

A body is dropped from a high tower and simultaneously another is projected up with a speed of
19.6 m/s from the base. Two seconds later they meet. The height of the tower is
1) 19.6m
2) 89.9m
3) 39.2m
4) none of these
​

Answer 1

Answer:19•6

Explanation:By using h=u^/2g

We get 19•6^/2(9•8)=19•6

Answer 2

Answer:option (3).39.2m.

Explanation:the question can be divided into two cases.

1st:(the downward distance that was travelled by the stone that was thrown)+

(The distance that was travelled by the stone thrown upwards).

Important note is that one has to first set a direction positive for the whole problem for thest type of questions.

So upward direction:

u=19.6 a=-10 m/s^2 t=2sec

By equating in s=ut+1/2a(t)^2

We get upward distance as (39.2m-20m).

The calculating the downward distance travelled the stone is

Since the stone is thrown with free fall it's u=0.

So,u=0, a=+10m/s^2, t=2 sec

Again equating in the equation s=ut +1/2a(t)^2.

We get downward distance is 20 m.

So by adding both upward and downward distances we get:

(39.2-20)+(20)=39.2m

(From ground)