A particle of mass 1 kg is fired with velocity 50m/s at an angle of 60...
Problem
A particle of mass 1 kg is fired with velocity 50m/s at an angle of 60degree from horizontal. It is acted by viscous force of 0.2v during it's journey. The horizontal distance traveled by it in first 10 seconds is:
- A) 90m
- B) 108m
- C) 125m
- D) 213m
Solution
Let's assume that the particle is moving in a viscous medium with a drag force of 0.2v, where v is the velocity of the particle. The horizontal and vertical components of the initial velocity are:
- Initial horizontal velocity, ux = 50 cos 60 = 25 m/s
- Initial vertical velocity, uy = 50 sin 60 = 43.3 m/s
Horizontal motion
The horizontal motion of the particle is not affected by the drag force, so the particle moves with a constant horizontal velocity of 25 m/s. The horizontal distance traveled by the particle in the first 10 seconds is:
dx = uxt = 25 x 10 = 250 m
Vertical motion
The vertical motion of the particle is affected by the drag force. The force acting on the particle in the vertical direction is:
Fy = -mg - bvy
where m is the mass of the particle, g is the acceleration due to gravity, b is the drag coefficient and vy is the velocity of the particle in the vertical direction.
The acceleration of the particle in the vertical direction is:
ay = (Fy/m) = (-mg - bvy)/m
Using the kinematic equation:
vy = uy + ayt
we can find the velocity of the particle in the vertical direction at any time t.
Using the kinematic equation:
dy = uyt + (1/2)ayt2
we can find the vertical distance traveled by the particle at any time t.
The time taken by the particle to reach the maximum height can be found by equating the velocity of the particle in the vertical direction to zero:
vy = 0 = uy + ayt
t = uy/ay
The maximum height reached by the particle can be found by substituting the time t in the vertical displacement