Problem:
From a point on the ground at a distance of 2 meters from the foot of a vertical wall, a ball is thrown at an angle of 45 degrees. The ball just clears the top of the wall and strikes the ground at a distance of 4 meters on the other side. Find the height of the wall.

Method of Calculation:
To find the height of the wall, we can break down the problem into two components: the horizontal and vertical motion of the ball.

Horizontal Motion:
The horizontal motion of the ball remains unaffected by the vertical motion. Since the ball strikes the ground at a distance of 4 meters on the other side, the horizontal distance traveled by the ball is 4 meters.

Vertical Motion:
The vertical motion of the ball can be analyzed separately. We know that the ball is thrown at an angle of 45 degrees and just clears the top of the wall. Let's denote the initial velocity of the ball as 'v' and the height of the wall as 'h'.

Using the concept of projectile motion, we can find the time taken by the ball to reach its maximum height and the time taken to reach the ground.

1. Maximum Height:
The time taken to reach the maximum height can be calculated using the formula:
t = (2v*sinθ)/g

Where:
t = time taken
v = initial velocity of the ball
θ = angle of projection
g = acceleration due to gravity

Since the ball just clears the top of the wall, the maximum height reached by the ball is equal to the height of the wall (h).

2. Time of Flight:
The total time of flight can be calculated using the formula:
T = (2v*sinθ)/g

Where:
T = total time of flight

3. Range:
The range of the projectile can be calculated using the formula:
R = v^2*sin(2θ)/g

Where:
R = range of the projectile

In this case, the range of the projectile is equal to 4 meters.

Solving the Equations:
1. Maximum Height:
Since the ball just clears the top of the wall, the maximum height reached is equal to the height of the wall (h). Therefore, we can write:
h = (v^2*sin^2θ)/(2g) ----(1)

2. Time of Flight:
The total time of flight can be written as:
T = (2v*sinθ)/g

3. Range:
The range of the projectile can be written as:
4 = v^2*sin(2θ)/g

Using the Value of T in the Range Equation:
Substituting the value of T from equation (2) into the range equation (3), we get:
4 = v^2*sin(2θ)/g
4 = v^2*(2sinθ*cosθ)/g
2 = 2sinθ*cosθ
sin(2θ) = 1
2θ = 90
θ = 45 degrees

Calculating the Maximum Height:
Using the value of θ in equation (1), we can calculate the maximum height (h):
h = (v^2*sin^2(45))/(2g)
h = (v^2*(1/2))/(2g)
h = (v^

1.33 m or 4/3 m.* Range= 6m (2+4)angle=45°according to the equationy(h)=tan (angle)x - gx^2 ÷ 2u^2 cos^(angle)h= 2-40/u^2 we know that R= u^2 sin2(angle)/g=6=: u^2 =60 substituting , h= 4/3