A particle A is projected vertically upwards from a point O on horizontal ground with speed 20ms. At the same instant a particle B is released from rest at point P which is 30 m above the ground. The point P Is not directly above O ignore air resistance. Find thr time that elapses between B arrival on the ground and A arrival on the ground.
Ii) the speeds of A and B at the instant when they are at the same height and show that A and B are moving in opposite directions at this instant.
To find the time elapsed between B's arrival on the ground and A's arrival on the ground, we first need to find the time of flight for each particle individually.
1) Time of flight for particle A:
The initial velocity of particle A is 20 m/s and it's projected vertically upwards. On its upward journey, it will eventually reach its highest point where its velocity becomes zero. At this highest point, its displacement is equal to the height of point P above the ground, which is 30 m. Using the equation of motion for displacement,
s = ut + (1/2)at^2
Where:
s = displacement (30 m)
u = initial velocity (20 m/s)
t = time
When the velocity becomes zero, the final displacement is 30 m. Rearranging the equation, we have:
30 = (20)t + (1/2)(-9.8)t^2
Simplifying the equation, we get:
4.9t^2 - 20t + 30 = 0
Solving this quadratic equation will give us the time taken by particle A to reach its highest point. However, since particle A will eventually return to the ground, we need to double this time to find the total time of flight. Let's call this time "t1".
2) Time of flight for particle B:
Particle B is released from rest at point P, which is 30 m above the ground. The equation of motion for the displacement of particle B from point P is:
s = ut + (1/2)at^2
Where:
s = displacement (30 m)
u = initial velocity (0 m/s)
t = time
In this case, the acceleration is due to gravity, which is -9.8 m/s^2. Plugging in the values, we get:
30 = (0)t + (1/2)(-9.8)t^2
Simplifying the equation, we have:
4.9t^2 = 30
Solving this equation will give us the time taken by particle B to reach the ground. Let's call this time "t2".
Now, to find the time that elapses between B's arrival on the ground and A's arrival on the ground, we need to subtract t2 from 2*t1 (since particle A takes 2*t1 to complete its full flight). Therefore,
Time elapsed = 2*t1 - t2
Once you solve the quadratic equations for t1 and t2, you can substitute them into the above equation to find the time elapsed between B's arrival on the ground and A's arrival on the ground.
Now, let's move on to the second part of the question:
To determine the speeds of particles A and B at the instant when they are at the same height, we need to find the height at which they coincide.
Let's assume that the height at which they coincide is h meters above the ground. At this point, both particles have reached the same height, so their respective displacements should be equal. Using the equation of motion for displacement,
s = ut + (1/2)at^2
Where:
s = displacement (h m)
u = initial velocity
For particle A:
Since it's projected upwards with an initial velocity of 20 m/s, its equation becomes:
h = (20)t1 + (1/2)(-9.8)t1^2
Similarly, for particle B:
Since it's released from rest at point P, its equation is:
h = (0)t2 + (1/2)(-9.8)t2^2
By equating these two equations, we can solve for the value of t1 and t2 when they are at the same height. Once we have these values, we can substitute them back into the respective equations to find the speeds of particles A and B at that instant.
Finally, to show that particles A and B are moving in opposite directions at this instant, we can compare their velocities. Particle A was initially projected upwards, so its velocity at any point during its flight will be positive. On the other hand, particle B was initially released from rest and is accelerating downwards due to gravity, so its velocity at any point during its fall will be negative. Therefore, since their velocities have opposite signs, particles A and B are moving in opposite directions at the instant when they are at the same height.