a body projected vertically up from ground crosses a point Pafter 4 seconds and again after 6 seconds Maximum height reached by that body above the ground is (g=10m/s)
Very badly answered the question
To find the maximum height reached by the body, we can use the equations of motion.
Let's assume the initial velocity of the body is u and the maximum height reached is h.
We are given that the body crosses point P after 4 seconds and again after 6 seconds, which means it takes a total of 6 seconds for the body to complete its upward and downward motion.
Using the equation of motion for displacement (s) with constant acceleration:
s = ut + (1/2)at^2
For the upward motion (from ground to maximum height), the time taken is 4 seconds:
0 = ut - (1/2)gt^2
For the downward motion (from maximum height to ground), the time taken is 2 seconds (6 seconds - 4 seconds):
h = 0 - (1/2)g(2)^2
Simplifying these equations, we have:
Equation 1: ut - 2g = 0
Equation 2: h = 2g
Now, we have two equations with two unknowns (u and h). Let's solve them simultaneously.
From Equation 2, we can find the value of g:
h = 2g
h = 2(10)
h = 20 meters
Therefore, the maximum height reached by the body above the ground is 20 meters.
To find the maximum height reached by the body, we can use the equations of motion for vertical motion.
Let's assume that the initial velocity of the body is u and the time it takes to reach maximum height is t.
1. The equation for the height reached at any time t is given by:
h = ut + (1/2)gt^2
Since the body is projected vertically up, the initial velocity u will be positive:
h1 = u(4) - (1/2)(10)(4)^2
2. The time taken to reach the maximum height can be found by dividing the total time taken to reach point P by 2:
t = (4 + 6) / 2 = 5 seconds
The equation for the height reached at time t will be:
h2 = u(5) - (1/2)(10)(5)^2
3. At the maximum height, the body will momentarily stop moving before it starts to fall downward. Therefore, the final velocity at this point will be zero.
v = u + gt
0 = u + (10)(t)
Solving for u, we get:
u = -10t
Substituting u into the equation for height at the maximum height:
h2 = (-10t)(5) - (1/2)(10)(5)^2
4. To find the maximum height, we need to equate h1 and h2 and solve for t:
u(4) - (1/2)(10)(4)^2 = (-10t)(5) - (1/2)(10)(5)^2
Solving this equation will give us the value of t, which corresponds to the maximum height reached by the body.
5. Finally, substitute the value of t into the equation for height at the maximum height to find the maximum height above the ground.
Note: It is important to convert the units as necessary to ensure all calculations are consistent. In this case, since the acceleration due to gravity is given as 10 m/s^2, all calculations are already in meters.
due to symmetry, the max height is reached after 5 seconds.
y = vi*t - 5t^2
at t=5, v=0, so since
v = vi - 10t,
vi-10*5=0
vi=50
y = 50t-5t^2
y(5) = 50*5-5*25 = 125