Question

hii friends !!

Plz solve my doubt

A person drops a stone from building of height 20m .At the same instant the front end of truck passes below the building moving constant acceleration of 1 m/s² velocity of 2m/s at that instant . lenght of truck if the stone just hit it's rear part is - ?

Answer 1

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given,
Heigth of the building = 20 m

Let the length of the truck be = L

The front end is supposed to travel this distance till the ston reaches the bottom,Thereforea=1 m/s^2

u'(initial velocity of the truck) = 2m/s

v=?L=?

H = 20the time taken by the stone

to reach the bottom :H = ut+1/2*g*t^2

20 =0*t+1/2*10*t^2

t^2=2*20/10 =2 sec

Use this time to get "L

"L=u't+ 1/2*a*t^2

L =2*2 + 1/2*1*2*2

L= length of the truck = 6 m

HOPE IT WILL HELP YOU

Answer 2

Answer:

~* THE ANSWER OF YOUR QUESTION ..*~

given,

Heigth of the building = 20 m

Let the length of the truck be = L

The front end is supposed to travel this distance till the ston reaches the bottom,Thereforea=1 m/s^2

u'(initial velocity of the truck) = 2m/s

v=?L=?

H = 20the time taken by the stone

to reach the bottom :H = ut+1/2*g*t^2

20 =0*t+1/2*10*t^2

t^2=2*20/10 =2 sec

Use this time to get "L

"L=u't+ 1/2*a*t^2

L =2*2 + 1/2*1*2*2

L= length of the truck = 6 m

HOPE IT WILL HELP YOU