Question

A stone is projected up with a velocity u and at the same time another is dropped from a height 2u when they will meet in air

Answer 1

Answer:

Explanation:A stone is projected up with a velocity u and at the same time another is dropped from a height 2u. When will they meet in air?

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these two stones say S1 and S2 are being thrown at the time t=0

but one from ground(S1) with initial velocity u m/s upward

and another being dropped from a height 2u meters

off course with initial velocity =0

imagine the gravitational force were switched off…..for a brief time…then S2 will remain there at height 2u m and S1 will reach it in 2 sec with a uniform speed u.

interesting so put on the acceleration due to gravity acting downward on each of them and not depending on the bodies,

so again it will meet after 2 sec. as the handicap of the two are similar…

.if the S2 gets speed up per sec by g then S1 will get its speed down by the same amount so the event of meeting together will be after the same time interval. calculate it.

after say a time t’ sec they meet…. the top one(s2)

would have fallen h= (1/2) .g . t’^2

and the down one (s1) must have gone up by a height

2u - h = u.t’ - (1/2).g. t’^2

h and (1/2) g.t’^2 will get cancelled

so 2u = u.t’ therefore time taken to meet t’ =2 sec.

Answer 2

For the above-given question, refer to the following given diagram.

Concept:

We need to utilize the second kinematical equation of motion. s = ut + 1/2gt²

Given:

Height, H = 2u

Find:

We need to determine the time at which the stones will meet in the air.

Solution:

The variables of motion are related to one another using kinematic equations.

Kinematics investigates the transfer of motion and energy among linked bodies. There is a two-way interaction in many of the interactions that have been mentioned, meaning that the projectile and the target are the two interacting parties.

We have, H = 2u

Suppose they meet at point B after time t.

Using the second kinematic equation of motion we have, s = ut + 1/2gt²

We have,

OB = ut - 1/2 gt² and AB = 1/2 gt²

Thus, OB + AB = H

H = ut - 1/2 gt² +  1/2 gt²

2u = ut

t = 2 seconds is the time

Thus, when a stone is projected up with a velocity of u and at the same time another is dropped from a height of 2u they will meet in the air at the time, t = 2 seconds.

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