a particle ia projected upward with a velocity of u m/s and after an interval of t(s) anoda particle is projected upward from the same point and wit the same initial velocity.proof dat they will meet at a height of (4u^2/8g-g^2t^2/8g)
To prove that the two particles will meet at a height of (4u^2/8g - g^2t^2/8g), let's break down the problem into steps.
Step 1: Determine the equations of motion for both particles.
We can use the equations of motion for uniformly accelerated motion:
For the first particle:
- Initial velocity: u m/s (projected upward)
- Acceleration: g m/s^2 (due to gravity acting downward)
- Final velocity: v (unknown)
- Displacement: S (unknown)
Using the equation v = u + at, where a = -g (negative because the particle is moving upward against the force of gravity), we can solve for v:
v = u - gt ----(Eq. 1)
Using the equation S = ut + (1/2)at^2, we can find the displacement:
S = ut - (1/2)gt^2 ----(Eq. 2)
For the second particle (projected after time t):
- Initial velocity: u m/s (projected upward)
- Acceleration: g m/s^2 (due to gravity acting downward)
- Final velocity: v' (unknown)
- Displacement: S' (unknown)
Using the same equations, the equations of motion for the second particle are:
v' = u - g(t + t') ----(Eq. 3)
S' = u(t + t') - (1/2)g(t + t')^2 ----(Eq. 4)
Step 2: Determine the time at which the two particles meet.
To find the time t' at which the two particles meet, we need to equate their displacements and solve for t':
S = S'
ut - (1/2)gt^2 = u(t + t') - (1/2)g(t + t')^2
Simplifying this equation, we get:
ut - (1/2)gt^2 = ut + ut' - (1/2)gt^2 - gtt' - (1/2)gt'^2
Canceling out similar terms, we get:
0 = ut' - (1/2)gt'^2
Factoring out t' from the equation, we have:
0 = t'(u - (1/2)gt')
t'(u - (1/2)gt') = 0
Since t' cannot be zero, we have:
u - (1/2)gt' = 0
Solving for t', we get:
t' = 2u/g ----(Eq. 5)
Step 3: Substitute t' into the equations of motion.
Now that we have the value of t', we can substitute it back into Eq. 3 and Eq. 4 to find the final velocity v' and displacement S'.
For Eq. 3:
v' = u - g(t + t')
v' = u - g(t + 2u/g)
v' = u - gt - 2u
v' = u - u - 2u
v' = -2u ----(Eq. 6)
For Eq. 4:
S' = u(t + t') - (1/2)g(t + t')^2
S' = u(t + 2u/g) - (1/2)g(t + 2u/g)^2
S' = ut + 2u^2/g - (1/2)g(t^2 + 4ut/g + 4u^2/g^2)
S' = ut + 2u^2/g - (1/2)gt^2 - 2ut - 2u^2/g - 2u^2/g^2
S' = -gt^2 - 2ut - (1/2)gt^2 - 2ut - 2u^2/g - 2u^2/g^2
S' = -2gt^2 - 4ut - 2u^2/g - 2u^2/g^2
S' = -2g(t^2 + 2ut + u^2/g + u^2/g^2)
S' = -2g(t + u/g)^2
Step 4: Substitute values into the equation for meeting height.
Now we have the expressions for v' and S'. The meeting height is given by the difference between the maximum height reached by the first particle and the displacement of the second particle at time t':
Meeting height = S - S'
= ut - (1/2)gt^2 - (-2g(t + u/g)^2)
= ut - (1/2)gt^2 + 2g(t + u/g)^2
= ut - (1/2)gt^2 + 2g(t^2 + 2t(u/g) + u^2/g^2)
= ut - (1/2)gt^2 + 2gt^2 + 4ut(u/g) + 2u^2/g
Simplifying and rearranging terms, we get:
Meeting height = -gt^2 + 4ut + 2u^2/g
= (4u^2 - g^2t^2)/8g
Therefore, the two particles will meet at a height of (4u^2/8g - g^2t^2/8g).