Problem:
A body projected upwards is at the same height from the ground at t=3s and t=7s. Find the maximum height attained by it.

Solution:
Let's assume that the initial velocity of the body is u and the acceleration due to gravity is g.

Step 1:
We know that the height of the body at any time t is given by the formula:

h = ut - 1/2 gt^2

Step 2:
At t=3s and t=7s, the height of the body will be the same. Therefore, we can write:

h1 = u*3 - 1/2*g*(3)^2
h2 = u*7 - 1/2*g*(7)^2
h1 = h2

Step 3:
Solving the above equation, we get:

u = 21/4 g
Substituting this value of u in the formula for h, we get:

h = 21/4 g (t - t^2/8)

Step 4:
To find the maximum height attained by the body, we need to find the time at which the height is maximum. For this, we can differentiate the above equation with respect to t and equate it to zero.

dh/dt = 21/4 g (1 - t/4) = 0
t = 4

Step 5:
Substituting t=4 in the equation for h, we get:

hmax = 21/4 g (4 - 4^2/8)
hmax = 21/8 g (4 - 2)

Step 6:
Simplifying the above equation, we get:

hmax = 21/8 g

Therefore, the maximum height attained by the body is 21/8 times the acceleration due to gravity.

Explanation:
When a body is projected upwards, it goes up and then comes down due to the force of gravity. The maximum height attained by the body is the highest point that it reaches during its upward journey. At this point, the velocity of the body becomes zero and it starts to come down. The time at which the body reaches the maximum height can be found by equating the derivative of the height equation to zero. Once we have the time, we can find the maximum height by substituting it in the height equation.

time of flight= (t1+t2 )/2where t1 and t2 are time at which it is at same heightT = (3+7)/2= 5so taking downward motion , as free fall h = gt²/2. (where T =2t)h = 10 *5*5/2 = 125 (if g= 10)if g= 9.8 h=122.5