IIT JEE , AIEEE Forum - Physics , Mechanics

rohit456

Q. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1m/s^2 and the projection velocity in the vertical direction is 9.8m/s. How far behind the boy will the ball fall on the car?

kirtana

hiii

ok u get the time required for the ball to reach the ground as 2secs by normal kinematic equations..

the distance travelled by the car in that time is given by

s = ut + 1/2 at²

s= 1/2 * 1*2*2= 2 meters

that's right

isnt it??

edison

Time taken by the ball to reach ground = 2 sec

when the ball was projected vertically upwards then its horizontal velocity = horizontal velocity of car at the moment when the ball was released = u (say)

so, horizontal distance travelled by the ball in 2 sec = ut = 2u meters ...... (1)

(Remember here there is no component of acceleration for the ball in horizontal direction once it is released)

Distance travelled by the car is given by

s = ut + (1/2)at²

= 2u + (1/2)*1*(2)²

or s =2u + 2 ............(2)

From (1) and (2)

The distance at which ball will fall behind the boy on the car is

= (2u + 2 ) - 2u = 2 meters

sinjan.j

I am going to give the solution to the problem......

I used frame of reference concept.

First, look at the diagram carefully.

Description of the diagram:

A is the position of the car when the ball is thrown straight upwards. The tall man is the person who throws and the short man is his friend.

Now, think from the friends point of view!

With respect to the friend the man is at rest. So, the man is not going to have horizontal velocity while throwing. So initial velocity of the car and the man is 0.

But the final velocity is not zero. So, the ball will not have any horizontal velocity with respect to the friend. It will only have vertical velocity.

Now look at diagram 2:

Now the car have moved with acceleration 1m/s² to the position B. The ball has gone on top and has come down on the ground.

So the net displacement of the ball is 0m.

Let us start our calculation now...!!!

Ball--------------------

s= ut - 1/2(gt² )

0= (9.8 x t) - 1/2(9.8 t² )

take 9.8t common on the RHS and we get,

(1-t/2)=0
therefore t= 2 sec.

Now ask yourself,

In the same time man has gone by how many metres?

well use the following formula:

s= ut + 1/2(at² )

and we get s= 2m.

So the car has traveled by 2metres.

Cheers!!!

BALGANESH

here is the easy method ,consider the relative motion of the bodies, since the boy on the car the horizantal speed of the ball will be same as that of the carat the instant of throwing .Now
v = u +at
0 = 9.8 -9.8t
t = 1
therefore total time taken to reach ground is 2 s
now since the car is accln at 1m/s2 , in 2s it would cover 2m

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