A car starts from rest and accelerates uniformly with $2 m / s^{2}$ . At t = 10 s a stone is dropped out of the window of the car which is 1 m in height from the ground. What will be the velocity and acceleration of the stone at t = 10 s? Neglect air resistance. (g = 10 m/s²)

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Solution

Velocity after 10 s:
Horizontal component: $v_{x} = u + a t = 0 + 2 \times 10 = 20 m / s$ (Horizontal)
Vertical component: $v_{y} = 0$ (since falling freely)
Net velocity: $v = v_{x}^{2} + v_{y}^{2} = 20^{2} + 0^{2} = 20 m / s$
Acceleration:
Horizontal direction: $a_{x} = 2 m / s^{2}$
Vertical direction: $a_{y} = 10 m / s^{2}$
Net acceleration: $a = a_{x}^{2} + a_{y}^{2} = 2^{2} + 10^{2} = 10.2 m / s^{2}$ .

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A car starts from rest and accelerates uniformly with 2m/s2. At t = 10 s a stone is dropped out of the window of the car which is 1 m in height from the ground. What will be the velocity and acceleration of the stone at t = 10 s? Neglect air resistance. (g = 10 m/s2)

Velocity after 10 s:Horizontal component: vx=u+at=0+2×10=20m/s (Horizontal)Vertical component: vy=0 (since falling freely)Net velocity: v=v2x+v2y=202+02=20m/sAcceleration:Horizontal direction: ax=2m/s2Vertical direction: ay=10m/s2Net acceleration: a=a2x+a2y=22+102=10.2m/s2.

A car starts from rest and accelerates uniformly with $2 m / s^{2}$ . At t = 10 s a stone is dropped out of the window of the car which is 1 m in height from the ground. What will be the velocity and acceleration of the stone at t = 10 s? Neglect air resistance. (g = 10 m/s²)