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 $\left.g=10\, ms ^{-2}\right)$

• A. $22.4\, ms ^{-1}$
• B. $25 \,ms ^{-1}$
• C. $27.2\, ms ^{-1}$
• D. $30.6\, ms ^{-1}$

Answer 1

Answer: (A)

Velocity of car (at $t=10\, s$ )
$=0+2 \times 10=20\, ms ^{-1}$
By the first law, the horizontal component of velocity is $20 ms ^{-1}$ throughout.
Vertical component of stone (at $t=11 s )$
$=0+(10 \times 1)=10\, ms ^{-1}$
$\Rightarrow $ Velocity of stone at $11^{\text {th }} sec =\sqrt{20^{2}+10^{2}}=\sqrt{500}$
$=22.4 \,ms ^{-1}$