A problem on relative motion from halliday resnick walker

[From: ] [author: ] [Date: 12-02-18] [Hit: ]

c)the velocity at the highest point of motion is zero.d)the velocity at the highest point of its motion is 5 m/s-It is assumed that the cart is moving along the positive direction of x-axis.To the observer on the ground (a) the initial components of velocities areUx = 4 m/sUy = 6 sin 30 = 6*(1/2) = 3 m/sUz = 6 cos 30 = 6*(√3/2) = 3√3 m/sResultant initial velocity = U = √(Ux² + Uy² + Uz²)=√(16 + 9 + 27) m/s = √52 m/s = 2√13 m/s----- option (b)(b) At the highest point, the velocity components are Vx = Ux = 4 m/sVy = Uy = 3 m/sVz = 0Resultant horizontal velocity at the highest point = V = √(Vx² + Vy²)= √(16 + 9) m/s = √25 m/s = 5 m/s .........

A cart is moving along a positive direction with a velocity of 4 m/s. a person on the cart throws a stone with a velocity of 6 m/s with respect to himself. in the frame of reference of the cart the stone iss thrown in the y=z plane making an angle of 30 degrees with the verticalz axis. then with respect to an observer on the ground

a) the initial velocity of the stone is 10 m/s

b) the initial velocity of the stone is 2 * root of 13 m/s.

c)the velocity at the highest point of motion is zero.

d)the velocity at the highest point of its motion is 5 m/s

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It is assumed that the cart is moving along the positive direction of x-axis.
To the observer on the ground
(a) the initial components of velocities are
Ux = 4 m/s
Uy = 6 sin 30 = 6*(1/2) = 3 m/s
Uz = 6 cos 30 = 6*(√3/2) = 3√3 m/s
Resultant initial velocity = U = √(Ux² + Uy² + Uz²)
=√(16 + 9 + 27) m/s = √52 m/s = 2√13 m/s ----- option (b)
(b) At the highest point, the velocity components are
Vx = Ux = 4 m/s
Vy = Uy = 3 m/s
Vz = 0
Resultant horizontal velocity at the highest point = V = √(Vx² + Vy²)
= √(16 + 9) m/s = √25 m/s = 5 m/s ....... option (d)

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