A cart 4.9m in length moving with a velocity of 4.9 m/sec is approaching a platf

Question

A cart 4.9m in length moving with a velocity of 4.9 m/sec is approaching a platform which is 9.8m above the plane of the cart's motion. You want to drop a particle onto the cart.
a. If you want to hit the center of the cart, how many seconds before the cart arrives under you must you drop the particle? How many meters before the cart arrives?

b. If you want to hit any place on the cart, during what time interval before the cart arrives must you drop the particle.

PLEASE INCLUDE A PICTURE IF POSSIBLE. THANKS =)

Explanation / Answer

a)so first calculate the time for the drop to fell down

h=9.8m

a=9.8m/s2

u=0m/sec

s=ut+.5at^2

9.8=0+.5*9.8t^2

=>2=t^2

=>t=1.414sec

so for the center to arrive to ur positon on seeing the platform the cart sould cover 4.9/2 m

i.e. 2.45m

time taen by cart to move 2.45m is

t1=2.45/4.9=.5sec

so no of seconds before the cart arrives under you must you drop the particle t2=t-t1

=1.414-.5

=.914sec

so the no of meters cart would have travelled is d=v*t2=4.9*.914

=4.48m

b)this only requires time for calculating the cart to travel the total length

the time taken by cart to travel the total length is 4.9/4.9=1 sec

and maximum time is when it should touch the front end and min time is when it should touchrear end

no of seconds before the cart arrives under you must you drop the particle to touch the front end is 1.414sec

no of seconds before the cart arrives under you must you drop the particle to touch the rear end is

1.414-1=.414sec

so time inteval is [.414,1.414]

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