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A truck starts from rest and accelerates uniformly at $2.0$ $\mathrm{ms}^{-2}$. At $t=10 \mathrm{~s}$, a stone is dropped by a person standing on the top of the truck ( $6 \mathrm{~m}$ high from the ground). What are the (a) velocity and (b) acceleration of the stone at $t=11 s$ ? (Neglect air resistance.)
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Given, $u=0, t=10 \mathrm{~s}, a=2 \mathrm{~ms}^{-2}$
The velocity of the truck when the stone is dropped, $v=u$ $+a t=0+2 \times 10=20 \mathrm{~ms}^{-1}$.
(a) When the stone is dropped, horizontal velocity, $v_x=v=20 \mathrm{~ms}^{-1} ; v_x$ remains constant as air resistance is neglected
In vertical direction, $a=g=9.8 \mathrm{~ms}^{-2}, u=0$, $t=11-10=1 \mathrm{~s}$.
$v_y=v=u+\mathrm{g} t=0+10 \times 1=10 \mathrm{~m} / \mathrm{s}$.
Resultant velocity,
$$
v=\sqrt{v_x^2+v_y^2}=\sqrt{20^2+10^2}=22.3 \mathrm{~ms}^{-1}
$$
Let $\theta$ be angle which $v_x$ makes with resultant velocity, $34^{\prime}$
(b) When the stone is dropped from the car, horizontal force on the stone is zero. Only acceleration of the stone $=$ acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ in the vertically downward direction. The path of the stone would be parabolic.
The velocity of the truck when the stone is dropped, $v=u$ $+a t=0+2 \times 10=20 \mathrm{~ms}^{-1}$.
(a) When the stone is dropped, horizontal velocity, $v_x=v=20 \mathrm{~ms}^{-1} ; v_x$ remains constant as air resistance is neglected
In vertical direction, $a=g=9.8 \mathrm{~ms}^{-2}, u=0$, $t=11-10=1 \mathrm{~s}$.
$v_y=v=u+\mathrm{g} t=0+10 \times 1=10 \mathrm{~m} / \mathrm{s}$.
Resultant velocity,
$$
v=\sqrt{v_x^2+v_y^2}=\sqrt{20^2+10^2}=22.3 \mathrm{~ms}^{-1}
$$
Let $\theta$ be angle which $v_x$ makes with resultant velocity, $34^{\prime}$
(b) When the stone is dropped from the car, horizontal force on the stone is zero. Only acceleration of the stone $=$ acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ in the vertically downward direction. The path of the stone would be parabolic.
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