Motion in a Straight Line Moderate Questions and Answers

D₄	=	0 + (a/2) × (2 × 4 - 1)	=	7
D₃	=	0 + (a/2) (2 × 3 - 1)	=	5

D₄	=	0 + (a/2) × (2 × 4 - 1)	=	7
D₃	=	0 + (a/2) (2 × 3 - 1)	=	5

In one dimensional motion, the body can have at a time one velocity but not two values of velocities.

In one dimensional motion, the body can have at a time one velocity but not two values of velocities.

Here, u = 0
We have, v² = u² + 2gh
⇒ v = √2gh = √2 × 10 × 20 = 20 m/s

Here, u = 0
We have, v² = u² + 2gh
⇒ v = √2gh = √2 × 10 × 20 = 20 m/s

No external force is acting, therefore, momentum is conserved.
By momentum conservation,
50 u + 0.5 × 2 = 0
where u is the velocity of man.

u = -	1	ms^-1
	50

Negative sign of u shows that man moves upward.
Time taken by the stone to reach the ground = 10/2 = 5s

Distance moved by the man 5 × (1/50) = 0.1 m
∴ when the stone reaches the floor, the distance of the man above floor = 10.1 m

No external force is acting, therefore, momentum is conserved.
By momentum conservation,
50 u + 0.5 × 2 = 0
where u is the velocity of man.

u = -	1	ms^-1
	50

Negative sign of u shows that man moves upward.
Time taken by the stone to reach the ground = 10/2 = 5s

Distance moved by the man 5 × (1/50) = 0.1 m
∴ when the stone reaches the floor, the distance of the man above floor = 10.1 m

For part AB
From 3rd equation of motion
v² = u² – 2gH

0 = u² – 2g(H/2) = u² – gH

H =	u²	=	10²	= 10 m
	g		10

For part AB
From 3rd equation of motion
v² = u² – 2gH

0 = u² – 2g(H/2) = u² – gH

H =	u²	=	10²	= 10 m
	g		10

Motion in a Straight Line