Question

A train moving along a straight line with constant acceleration a . A boy standing in the train throws a ball with a speed of 10m/s at 60 degree to horizontal the boy has to move 1.15m inside train to catch the ball at same initial height the accl of train is?

Answer 1

inside the accelerating train we add a fictitious force which acts against the direction of the acceleration

Answer 2

this fictitious force will be m - mass of the ball times a - acceleration of the train ..

do you think you can do it now ?

Answer 3

wat is fictitious force

Answer 4

newton's second law (and first) are valid inside inertial frame of reference ..
an accelerating train is not an inertial frame of reference .. in order to make those laws valid inside such frames we add fictitious forces ..

Answer 5

when you sit in an accelerating car and you have something hanging on your mirror you see it go backwards .. for you - you say ah! some force pushing it backwards ..
for someone who is not inside the car and watches the car from out side
he knows that the car accelerates hence something has to accelerate the object
so the string that the object hangs on it has to stretch so the object will accelerate with the car

Answer 6

u mean Pseudo Force...)

Answer 7

yes!

Answer 8

sorry if you dont use "fictitious force" in english

Answer 9

so you think you can do it ? or you need some more ?

Answer 10

i need help...) cant proceed

Answer 11

suppose train is at rest,,then FBD of ball is..
(assuming train was going rightwards initially with accn a)

now x(horizontal disp) = 1.15
a_x = a
u_x = 10 cos60

y(vertical disp)=0
u_y=10 sin60
a_y = -g

from vertical motion, solve for time
apply 2nd eqn of motion for horizontal and solve for a..

Answer 12

ok so we have initial velocity in x direction 10 cos60 = 5

so x = 5t - at^2/2
(a is the same acceleration of the train!)
1.15 = 5t - at^2 / 2

no as for y :
h = h + (10sin60) * t - gt^2/2
(10sin60) * t - gt^2/2 = 0
solve for t and plug it into
1.15 = 5t - at^2 / 2
then find a

Answer 13

@Coolsector Gravity..has to play a role here..)

Answer 14

we have gravity ..
"(10sin60) * t - gt^2/2 = 0" there is g here

Answer 15

T=1.73s

Answer 16

@shubhamsrg if the x component of the velocity of the ball is in the direction of the acceleration (and i think it is .. as i understood the problem) the ma has to be in the other direction

Answer 17

now plug it in here : 1.15 = 5t - at^2 / 2 and fine a

Answer 18

Yup...got...it..)

Answer 19

a=~5m/s^2