A truck starts from rest and accelerates uniformly at 2.0 m s^(-2).At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ?(Neglect air resistance.)

A truck starts from rest and accelerates uniformly at 2.0 m s^(-2).At

1.	A truck starts from rest and accelerates uniformly at 2.0 m s^(-2).At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ?(Neglect air resistance.)
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(a)Velocity of car ( at t = 10 s ) =` <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> + 2 x× 10 = 20 m s^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/>By the First Law, the <a href="https://interviewquestions.tuteehub.com/tag/horizontal-1029056" style="font-weight:bold;" target="_blank" title="Click to know more about HORIZONTAL">HORIZONTAL</a> component of velocity is` 20 m s^(-1)`throughout. Vertical component of velocity (at t = 11s) ` = 0 + 10 xx 1 = 10 ms^(-1)`<br/>Velocity of stone (at t = 11s) ` = sqrt( 20^2 + 10^2) = sqrt 500 = 22.4 ms^(-1)` at an angle of `tan^(-1) (1//2)` with the horizontal. (b) `10 ms^(-2)` vertically <a href="https://interviewquestions.tuteehub.com/tag/downwards-959173" style="font-weight:bold;" target="_blank" title="Click to know more about DOWNWARDS">DOWNWARDS</a> .</body></html>