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

Question

A truck starts from rest and accelerates uniformly at 2.0 ms -2. At t=10 s, a stone is dropped by a person standing on top of the truck (6 m high from the ground). What is the velocity of the stone at t=11 s? (Take .g=10 ms -2) (A) 22.4 ms -1 (B) 25 ms -1 (C) 27.2 ms -1 (D) 30.6 ms -1

Tardigrade Admin · Accepted Answer

Velocity of car (at t=10 s ) =0+2 × 10=20 ms -1 By the first law, the horizontal component of velocity is 20 ms -1 throughout. Vertical component of stone (at t=11 s ) =0+(10 × 1)=10 ms -1 ⇒ Velocity of stone at 11 text th sec =√202+102=√500 =22.4 ms -1

Q. 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 top of the truck $(6 m$ high from the ground). What is the velocity of the stone at $t = 11$ s? (Take $g = 10 m s^{- 2})$

$22.4 m s^{- 1}$

$25 m s^{- 1}$

$27.2 m s^{- 1}$

$30.6 m s^{- 1}$

Q. A truck starts from rest and accelerates uniformly at 2.0ms−2. At t=10s, a stone is dropped by a person standing on top of the truck (6m high from the ground). What is the velocity of the stone at t=11 s? (Take g=10ms−2)

22.4ms−1

25ms−1

27.2ms−1

30.6ms−1

Velocity of car (at t=10s ) =0+2×10=20ms−1 By the first law, the horizontal component of velocity is 20ms−1 throughout. Vertical component of stone (at t=11s) =0+(10×1)=10ms−1 ⇒ Velocity of stone at 11th sec=202+102​=500​ =22.4ms−1

Q. 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 top of the truck $(6 m$ high from the ground). What is the velocity of the stone at $t = 11$ s? (Take $g = 10 m s^{- 2})$

$22.4 m s^{- 1}$

$25 m s^{- 1}$

$27.2 m s^{- 1}$

$30.6 m s^{- 1}$