Answer:
To solve this problem, we can use the equations of motion for a particle moving in the vertical direction with constant acceleration g. The equation for the height (h) of the particle as a function of time (t) is:
h = ut - 1/2 gt^2
The particle will reach its maximum height when it has zero velocity, which occurs at the peak of its trajectory. The time taken to reach the peak can be found by setting the velocity equal to zero:
0 = u - gt
t = u/g
Substituting this value of t into the equation for h, we get:
h = u(u/g) - 1/2 g(u/g)^2
h = u^2/2g - u^2/2g
h = 0
This tells us that the particle reaches a height of zero at the end of its trajectory, which is not the maximum height. To find the maximum height, we need to consider the distance s between the point of projection and the wall.
Let t1 be the time taken by the particle to reach the wall, and t2 be the time taken to return to the ground. These times can be found by solving the quadratic equation:
s = ut1 + 1/2 gt1^2
0 = ut2 - 1/2 gt2^2
Using the quadratic formula, we get:
t1 = (u + sqrt(u^2 + 2gs))/g
t2 = (u - sqrt(u^2 + 2gs))/g
The maximum height is reached halfway between t1 and t2, which occurs at:
t = (t1 + t2)/2
t = (u/g)
Substituting this value of t into the equation for h, we get:
h = u(u/g) - 1/2 g(u/g)^2 - s
h = u^2/2g - gs/2
Therefore, the greatest height that a particle with initial velocity u can reach on a vertical wall at a distance s from the point of projection is u²/2g - gs²/2u².