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  • 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 of the stone at t = 11s ? (Neglect air resistance.

Q5.11 A truck starts from rest and accelerates uniformly at 2.0ms^{-2} . At t = 10s, 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 of the stone at t = 11s? (Neglect air resistance.)

Answers (1)

best_answer

The initial velocity of the truck is given as zero.

We need to find the final velocity (at t  =  10 s), so we will use the equation of motion  :

                                               v\ =\ u\ +\ at

or                                                  =\ 0\ +\ (2)10\ =\ 20\ m/s 

This is the velocity imparted to stone by the truck that's why it is a horizontal component of velocity.

The stone is dropped at t = 10 sec. so it has travelled 1 sec in the air (11 - 10 = 1 s). We need to find the final vertical velocity.

                                               v\ =\ u\ +\ at

                                                    =\ 0\ +\ 10(1)\ =\ 10\ m/s

Thus resultant of both the component is the required velocity.

                                           R\ =\ \sqrt{v_h^2\ +\ v_v^2}

or                                               =\ \sqrt{20^2\ +\ 10^2}

or                                               =\ 22.36\ m/s

Direction : 

                                   \tan \Theta \ =\ \frac{v_v}{v_h}\ =\ \frac{10}{20}

or                                       \Theta \ =\ 26.57^{\circ}

Posted by

Devendra Khairwa

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